Help in simplifying combinatorial sum $\frac{n!}{(n-k)!}-{1\over(n-k)!}{\sum _{m=1}^{k-1} (-1)^{m+1} (n-m)! S(k,k-m)}$

Notice that your expression can be stated as $$\sum _{m=0}^k(-1)^m\frac{(n-m)!}{(n-k)!}S(k,k-m)=\sum _{m=0}^k(-1)^{k-m}\frac{(n-k+m)!}{(n-k)!}S(k,m).$$
Now recall that $a^{\underline{b}}=\frac{a!}{(a-b)!}=a(a-1)\cdots (a-b+1)$ and $a^{\overline{b}}=a(a+1)\cdots (a+b-1)$ hence $a^{\underline{b}}=(-1)^b(-a)^{\overline{b}}$ and $(a)^{\underline{b}}=(a-b+1)^{\overline{b}}$

Now, notice that $$x^k=\sum _{m=0}^kx^{\underline{m}}S(k,m),$$ this can be understood as splitting the functions into the cardinal of their images( count how many functions from $[n]$ to $[x]$ there are with image having exactly $m$ elements).

$$\sum _{m=0}^k(-1)^{k-m}\frac{(n-k+m)!}{(n-k)!}S(k,m)$$ $$=\sum _{m=0}^k(-1)^{k-m}(n-k+m)^{\underline{m}}S(k,m)$$ $$=\sum _{m=0}^k(-1)^{k-m}(n-k+1)^{\overline{m}}S(k,m)$$ $$=\sum _{m=0}^k(-1)^{k}(-n+k-1)^{\underline{m}}S(k,m)$$ $$=(-1)^k(-n+k-1)^k.$$

In seeking to show that

$$\frac{n!}{(n-k)!}-\frac{1}{(n-k)!} \sum_{m=1}^{k-1} (-1)^{m+1} (n-m)! {k\brace k-m} = (n-k+1)^k$$

we follow the observation by @Phicar and simplify the LHS as follows:

$$\frac{n!}{(n-k)!}+\frac{1}{(n-k)!} \sum_{m=1}^{k-1} (-1)^{m} (n-m)! {k\brace k-m} \\ = \frac{n!}{(n-k)!}+\frac{1}{(n-k)!} \sum_{m=1}^{k} (-1)^{m} (n-m)! {k\brace k-m} \\ = \frac{1}{(n-k)!} \sum_{m=0}^{k} (-1)^{m} (n-m)! {k\brace k-m}.$$

We have using standard EGFs

$$\frac{1}{(n-k)!} k! [z^k] \sum_{m=0}^{k} (-1)^{m} (n-m)! \frac{(\exp(z)-1)^{k-m}}{(k-m)!} \\ = k! [z^k] \sum_{m=0}^{k} (-1)^{m} {n-m\choose k-m} (\exp(z)-1)^{k-m} \\ = k! [z^k] \sum_{m=0}^{k} (-1)^{k-m} {n-k+m\choose m} (\exp(z)-1)^{m}.$$

Now since $\exp(z)-1 = z +\cdots$ the coefficient extractor enforces the range and we may write

$$k! [z^k] (-1)^{k} \sum_{m\ge 0} (-1)^{m} {n-k+m\choose m} (\exp(z)-1)^{m} \\ = k! [z^k] (-1)^{k} \frac{1}{(1+\exp(z)-1)^{n-k+1}} \\ = k! [z^k] (-1)^{k} \exp(-(n-k+1)z).$$

We have at last

$$\bbox[5px,border:2px solid #00A000]{ (n-k+1)^k.}$$