show that $(k!)^{k^n+ k^{n-1} ~+k^{~n-2}~~+\cdots+k+1}|(k^{n+1})!$ Let $k$ is  postive integers,and $n\ge 0,n\in Z$,show that
$$(k!)^{k^n+ k^{n-1} ~+k^{~n-2}~~+\cdots+k+1}|(k^{n+1})!$$
if $n=0$,it is obvious 
$$(k!)^{k^n+\cdots+k+1}=k!|k!=(k^1)!$$

We just need to take a prime $p$ and show that:
$v_p( (k!)^{1+k+k^2+\dots + k^n})\leq v_p(k^{n+1}!)$
Or in other words:
$(1+k+k^2+\dots + k^n)v_p(k!)=\frac{k^{n+1}-1}{k-1}v_p(k!)\leq v_p(k^{n+1}!)$
we now apply the following theorem:
$v_p(m!)=\frac{m-s_p(m)}{p-1}$, where $s_p(m)$ is the sum of the digits of $m$ in base $p$.
So all we have to prove is:
$\frac{k^{n+1}-1}{k-1}(k-s_p(k))\leq k^{n+1}-S_p(k^{n+1})\iff k^{n+2}-k^{n+1}S_p(k)-k+s_p(k)\leq k^{n+2}-k^{n+1}-ks_p(k^{n+1})+s_p(k^{n+1})\iff (k-1)s_p(k^{n+1})\leq (k^{n+1}-1)s_p(k)+k$.
This last inequality is clear,  just notice $s_p(k^{n+1})\leq k^ns_p(k)$
