Prove:
$v_{p}(p^k!) = 1+p+p^2+...+p^{k-1}$
Proving: Use legendre formula: $v_p (n!) = \sum_{i=1}^{\infty}\frac{n}{p^i}$. So $v_p (p^k!) = \sum_{i=1}^{\infty}\frac{p^k}{p^i} = \sum_{i=1}^{\infty}\ p^{k-i} = \frac{p^k}{p-1}$ where $|p| >1$.
As for me it's not what I need. Thanks for any help.
I'm not sure this is what you need. The exact form of legendre formula is
$$ v_p(n!)=\sum_{i=1}^{\infty}\Big\lfloor\frac n{p^i}\Big\rfloor. $$
For $n=p^k$, the terms where $i>k$ are clearly zero. When $i\leqslant k$, $n/p^i$ is an integer, and the desired result follows.