A proof of $\sum_{i=0}^{j}(-1)^i{j \choose i}(j^n-i^n)^z=j!$ I'm quite confused about how can this be possible in this sum.
let $j=nz$ and the sum 
$$\sum_{i=0}^{j}(-1)^i{j \choose i}(j^n-i^n)^z=j!$$
I would like to see proof of this sum, thank you.
 A: We seek to show that (here $j=nq$)
$$\sum_{k=0}^{j} (-1)^{k+j-q}
{j\choose k} (j^n - k^n)^q = j!$$
The LHS is
$$\sum_{k=0}^{j} (-1)^{k+j-q} {j\choose k}
\sum_{p=0}^q {q\choose p} j^{np} (-1)^{q-p} k^{j-np}
\\ = (-1)^{j} \sum_{p=0}^q {q\choose p} j^{np} (-1)^{p}
\sum_{k=0}^{j} (-1)^{k} {j\choose k} k^{j-np}.$$
Continuing,            
$$(-1)^{j} \sum_{p=0}^q {q\choose p} j^{np} (-1)^{p}
\sum_{k=0}^{j} (-1)^{k} {j\choose k} (j-np)!
[z^{j-np}] \exp(kz)    
\\ = (-1)^{j} \sum_{p=0}^q {q\choose p} j^{np} (-1)^{p}
(j-np)! [z^{j-np}] (1-\exp(z))^{j}.$$
Now $1-\exp(z) = - z - \cdots$ and we have $[z^{j-np}] (1-\exp(z))^{j}
= 0$ when $1\le p\le q,$ hence
$$(-1)^{j} \sum_{p=0}^q {q\choose p} j^{np} (-1)^{p}
[[p=0]] (j-np)! [z^{j-np}] (1-\exp(z))^{j}
\\ = (-1)^{j} \sum_{p=0}^q [[p=0]] (j-np)! [z^{j-np}]
{q\choose p} j^{np} (-1)^{p}
(1-\exp(z))^{j}
\\ = (-1)^{j} j! [z^{j}] (1-\exp(z))^{j}
= j!$$
This is the claim.
A: This does not appear correct.
If $z=0$
the left side is zero and
the right side is
$j!$.
Also,
$z$ and $n$ should be in the result.
