Deriving the Normalization formula for Associated Legendre functions: Stage $1$ of $4$
We show the following is valid for $0\leq m\leq L$ \begin{align*} \frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L =\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L \end{align*}
We apply the Leibniz Product Rule \begin{align*} (f\cdot g)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)} \end{align*} to $(x^2-1)^L=(x+1)^L(x-1)^L$ according to the hint.
We obtain \begin{align*} &\frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L\\ &\quad=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L\tag{1}\\ &\quad=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{2}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=0}^{L-m}\frac{(L+m)!}{j!(L-m-j)!}\cdot\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{3}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=m}^{L}\frac{(L+m)!}{(j-m)!(L-j)!}\cdot\frac{L!}{(L+m-j)!}(x+1)^{L+m-j}\frac{L!}{j!}(x-1)^{j}\tag{4}\\ &\quad=\frac{(L-m)!}{(L+m)!}(x^2-1)^m \sum_{j=m}^{L}\binom{L+m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(j-m)!}(x-1)^{j-m}\tag{5}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=m}^{L}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{6}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L+m}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{7}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L\tag{8}\\ \end{align*} and the claim is finished.
Comment:
In (1) we apply the Leibniz product rule.
In (2) we do the differentiation which is not too hard since we have powers of linear factors only \begin{align} \frac{d^j}{dx^j}(x+1)^L &= L(L-1)(L-2)\cdots(L-j+1)(x+1)^{L-j}\\ &=\frac{L!}{(L-j)!}(x+1)^{L-j}\\ \frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L &= \frac{L!}{(L-(L-m-j))!}(x-1)^{L-(L-m-j)}\\ &=\frac{L!}{(m+j)!}(x-1)^{m+j} \end{align}
In (3) we factor out $(L-m)!$ from $\binom{L-m}{j}$ and expand the expression with $\frac{(L+m)!}{(L+m)!}$.
In (4) we shift the index $j$ to start from $m$ and substitute in the expression $j\rightarrow j-m$ accordingly.
In (5) we factor out $(x+1)^m(x-1)^m=(x^2-1)^m$ and rearrange the factorials.
In (6) we write the expression using derivatives.
In (7) we extend the range of $j$ to $0\leq j \leq L+m$ without changing anything, since we are just adding zeros.
More detailed: Since $(x+1)^L$ and $(x-1)^L$ are polynomials in $x$ of degree $L$ we get \begin{align*} \frac{d^j}{dx^j}(x+1)^L&=0\qquad\qquad L<j\leq L+m\\ \frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L&=0\qquad\qquad 0\leq j<m\\ \end{align*}
- In (8) we use Leibniz' rule again.
Here's a hint: evaluating $$\frac{\mathrm{d}^{L-(m+2)}}{\mathrm{d}x^{L-(m+2)}}(x+1)^L$$ is easier than you think.
You can figure out what happens when you differentiate $f(t)=t^L$ repeatedly, right? You get $f'(t)=L t^{L-1}$, $f''(t)=L(L-1) t^{L-2}$, $f'''(t)=L(L-1)(L-2)t^{L-3}$, and so on. And $L(L-1)(L-2)$ can be written as $L!/(L-3)!$ if you prefer that, and so on. Now just write down what the pattern would be if you did it $L$ times, then replace $t$ by $x+1$ and $L$ by $L-(m+2)$.
Note: Please find here a former variation of my current answer. OP was asking to republish it to make the connection with some of OPs comments and his related question plausible.
Both answers are more or less the same. Originally I've decided to apportion it into two somewhat smaller pieces to enhance readability. The current answer was only written since OP was explicitely asking for a one sided approach.
Here we transform the left hand side to get a convenient representation. Then we transform the right hand side to obtain the same representation and equality is shown.
The following is valid for $0\leq m\leq L$ \begin{align*} \frac{d^{L-m}}{dx^{L-m}}\left(x^2-1\right)^L =\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L \end{align*}
We apply the Leibniz Product rule \begin{align*} (f\cdot g)^{(n)}=\sum_{j=0}^n\binom{n}{j}f^{(j)}g^{(n-j)} \end{align*} according to the hint to $(x^2-1)^L=(x+1)^L(x-1)^L$.
We obtain from the LHS \begin{align*} \frac{d^{L-m}}{dx^{L-m}}&\left(x^2-1\right)^L\\ &=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L-m-j}}{dx^{L-m-j}}(x-1)^L\tag{1}\\ &=\sum_{j=0}^{L-m}\binom{L-m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(m+j)!}(x-1)^{m+j}\tag{2}\\ &=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{3} \end{align*}
Comment:
In (1) we apply the Leibniz product rule
In (2) we do the differentiation which is not too hard since we have powers of linear factors only \begin{align*} \frac{d^j}{dx^j}(x+1)^L&=L(L-1)(L-2)\cdots(L-j+1)(x+1)^{L-j}\\ &=\frac{L!}{(L-j)!}(x+1)^{L-j} \end{align*}
In (3) we factor out $(L-m)!$ from $\binom{L-m}{j}$ and rearrange the factorials to get nice binomial coefficients.
And now the RHS \begin{align*} &\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m\frac{d^{L+m}}{dx^{L+m}}\left(x^2-1\right)^L\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L+m}\binom{L+m}{j}\frac{d^j}{dx^j}(x+1)^L\frac{d^{L+m-j}}{dx^{L+m-j}}(x-1)^L\tag{1}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=m}^L\binom{L+m}{j}\frac{L!}{(L-j)!}(x+1)^{L-j}\frac{L!}{(j-m)}!(x-1)^{j-m}\tag{5}\\ &\quad=\frac{(L-m)!}{(L+m)!}\left(x^2-1\right)^m \sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j-m}\frac{L!}{j!}(x-1)^j\tag{6}\\ &\quad=\frac{(L-m)!}{(L+m)!} \sum_{j=0}^{L-m}\binom{L+m}{j+m}\frac{L!}{(L-j-m)!}(x+1)^{L-j}\frac{L!}{j!}(x-1)^{j+m}\tag{7}\\ &\quad=(L-m)!\sum_{j=0}^{L-m}\binom{L}{j}\binom{L}{m+j}(x+1)^{L-j}(x-1)^{m+j}\tag{8} \end{align*} and the claim follows.
Comment:
In (5) we do it similarly to (2). Note, that we also shrink the range of the index variable to $m\leq j\leq L$, since for all other values of $j$ the differentiation give zero.
In (6) we shift the index $j$ to start from $0$ and substitute in the expression $j\rightarrow j+m$
In (7) we apportion $(x^2-1)^m=(x+1)^m(x-1)^m$
In (8) we cancel $(L+m)!$ and reorder the factorials similarly to (3)