Can't recognize the series. Can somebody take a look? I am solving a question as part of which I got the below mentioned series. I tried a lot but couldn't recognize this.
$$
^m C_m m^n -  {^m C_{m-1} (m - 1)^n} + \cdots \pm {^mC_1} 1^n
$$
I am sure it's expansion of some famous series.
Can somebody help?
 A: If you define $(Df)(x)=f(x+1)-f(x)$ then your expression is 
$$
(D^m f)(x)=\binom{m}{m} f(x+m)-\binom{m}{m-1}f(x+m-1)\pm…+(-1)^{m-1}\binom{m}1f(x+1)+(-1)^m\binom{m}{0}f(x+0)
$$ 
where $f(x) = x^n$ and evaluated at $x=0$, in total it is an $m$-th derivative approximated by an $m$-th order difference quotient of step size 1.



*

*If $f$ is a polynomial of degree $n$, then $D^mf$ is a polynomial of degree $n-m$ if that is non-negative, or else the zero polynomial.

*If $n<m$ and $f(x)=x^n$, then $(D^mf)(x)=0$.

*If $n=m$ and $f(x)=x^n$, then $(D^mf)(x)=m!$.

A: We use the notation $\binom{m}{j}$  instead of $^mC_j$ and we also use the coefficient extraction operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.
\begin{align*}
  n![z^n]e^{jz}=j^n
  \end{align*}

We obtain  for   $m\geq 1$
  \begin{align*}
\sum_{j=1}^m\binom{m}{j}(-1)^{m-j}j^n
&=\sum_{j=1}^m\binom{m}{j}(-1)^{m-j}n![z^n]e^{jz}\\
&=n![z^n]\sum_{j=1}^m\binom{m}{j}\left(e^z\right)^j(-1)^{m-j}\\
&=n![z^n]\left((e^z-1)^m-1\right)\\
&=n![z^n]\left(\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)^m-1\right)\\
&=\begin{cases}
-1&\qquad n=0\\
0&\qquad 1\leq n < m\\
m!&\qquad n=m\\
n![z^n](e^z-1)^m&\qquad n\geq m
\end{cases}
\end{align*}

