TITLE: Prove $\sum\limits_{j=1}^k(-1)^{k-j}j^k\binom{k}{j}=k!$
QUESTION [21 upvotes]: In reading about A polarization identity for multilinear maps by Erik G F  Thomas, I am led to prove the following combinatorial identity, which I cannot find anywhere, nor do I have any idea how to prove it.
$$\sum\limits_{j=1}^k(-1)^{k-j}j^k\binom{k}{j}=k!.$$
After some reductions (including $\binom{k+1}{j}=\binom{k}{j}+\binom{k}{j-1}$) and "simplifications", I am still stuck and have no clues how to proceed.
I do not know if this is true, either, though I think it should be.

Any help or reference is sincerely welcomed, thanks in advance.

REPLY [3 votes]: Here is a variant based upon the coefficient of operator $[z^k]$ used to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
[z^j](1+z)^k=\binom{k}{j}\qquad\qquad\text{or}\qquad\qquad k![z^k]e^{jz}=j^k
\end{align*}

We obtain
  \begin{align*}
\sum_{j=1}^k(-1)^{k-j}j^k\binom{k}{j}
&=\sum_{j=0}^\infty(-1)^{k-j}k![z^k]e^{jz}[u^k](1+u)^k\tag{1}\\
&=(-1)^kk![z^k]\sum_{j=0}^{\infty}\left(-e^{z}\right)^j[u^k](1+u)^k\tag{2}\\
&=(-1)^kk![z^k](1-e^z)^k\tag{3}\\
&=k![z^k]\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)^k\tag{4}\\
&=k!
\end{align*}
  and the claim follows.

Comment:

In (1) we apply the coefficient of operator twice and extend the lower and upper limit of the series without changing anything, since we add only zeros.
In (2) we use the linearity of the coefficient of operator and do some rearrangements.
In (3) we use the substitution rule with $u=-e^z$:

\begin{align*}
A(z)=\sum_{j=0}^\infty a_jz^j=\sum_{j=0}^\infty z^j [u^k]A(u)
\end{align*}

In (4) we see that in each of the $k$ factors  $(e^z-1)$ only $z^1$ provide a contribution to $z^k$.