A nasty indefinite integral $ \int \frac{1}{x(x+1)(x+2)(x+3)(x+4) ... (x+m)} dx $

A simpler approach is possible without all the other preliminary work. Let $$q_m(x) = \prod_{k=0}^m (x+k), \quad f_m(x) = \frac{1}{q_m(x)}.$$ Then $f$ admits a partial fraction decomposition of the form $$f_m(x) = \sum_{n=0}^m \frac{A_n}{x+n} \tag{1}$$ for suitable constants $A_0, \ldots, A_m$ which we wish to find, hence an antiderivative of $f$ is $$\int f_m(x) \, dx = \sum_{n=0}^m A_n \log |x+n|. \tag{2}$$ (I have omitted the constant of integration for convenience.) So all that remains is to determine the form of $A_n$. To do this, we observe that $$1 = q_m(x) \sum_{n=0}^m \frac{A_n}{x+n} = \sum_{n=0}^m p_n(x) A_n,$$ where $$p_n(x) = \prod_{k \ne n} (x+k) = (-1)^n \prod_{k=0}^{n-1} (-x-k) \prod_{k=n+1}^m (k + x).$$ Then in particular $$p_n(-n) = (-1)^n \prod_{k=0}^{n-1} (n-k) \prod_{k=n+1}^m (k-n) = (-1)^n n!(m-n)! = \frac{(-1)^n m!}{\binom{m}{n}},$$ and $p_n(-k) = 0$ for all other nonnegative integers $k \le m$ not equal to $n$. Therefore, $$A_n = \frac{1}{p_n(-n)} = \frac{(-1)^n}{m!} \binom{m}{n}$$ and $$\int f_m(x) \, dx = \sum_{n=0}^m \frac{(-1)^n}{m!} \binom{m}{n} \log |x+n| + C$$ as claimed.

At this line: "Now let's consider the first term. We set $x=0$ and we get:" $$1 = {C_0}(0+1)(0+2) \cdots (x+m) = m!C_0$$ When you set $x=0$ there won't be any $x$ left.

For $C_2$ there is no mistake but then for the general term you did the same typo.

You can't have after you set $x=-m$: $$1 = C_m(-m)(-m+1)(-m+2) \cdots (\color{red}x+m -1)$$ And for the last part for some reason, you have $C_k$ in the denominator. $$\begin{align*} 1 &= C_k(-k)(-k+1)(-k+2) \cdots (-1) (1)(2) \cdots (-k + m - 1) \\ 1 &= (-1)^k C_k(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) \\ C_k &= \frac{ 1 }{ (-1)^k \color{red}{C_k}(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) } \\ C_k &= \frac{ k! }{ (-1)^k \color{red}{C_k}(k-1)(k-2) \cdots (1)m! } \\ C_k &= \frac{ \binom {m}{k} }{ (-1)^k m! } \\ \end{align*}$$ Other than this everything it's correct.

We could set up a difference equation. Partial fractions indicates that $$f_m(x)=\frac1{\prod_{k=0}^m(x+k)}=\sum_{k=0}^m\frac{A_k^{(m)}}{x+m}$$ And it is a simple calculation to show that $$f_{m-1}(x)-f_{m-1}(x+1)=mf_m(x)$$ So, comparing coefficients of $\frac1{x+k}$ we have $A_0^{(m-1)}=mA_0^{(m)}$, $A_{m-1}^{(m-1)}=-mA_m^{(m)}$, and $A_k^{(m-1)}-A_{k-1}^{(m-1)}=mA_k^{(m)}$ for $1\le k\le m-1$. If we let $A_k^{(m)}=\frac{(-1)^k}{m!}B_k^{(m)}$ then our difference equations read $B_0^{(m-1)}=B_0^{(m)}=\cdots=B_0^{(0)}=A_0^{(0)}=1$, $B_{m-1}^{(m-1)}=B_m^{(m)}=\cdots=B_0^{(0)}=1$ and $B_k^{(m-1)}+B_{k-1}^{(m-1)}=B_k^{(m)}$ for $1\le k\le m-1$. We recognize these as the difference equations for Pascal's triangle, so $B_k^{(m)}={m\choose k}$ so it follows that $A_k^{(m)}=\frac{(-1)^k}{m!}{m\choose k}$ and $$\int f_m(x)dx=\sum_{k=0}^m\frac{(-1)^k}{m!}{m\choose k}\ln|x+k|+C$$