There is a problem which i tried a lot but didn't reach to any conclusions:How can i show that the highest power of n which is contained in $(n^r-1)!$ is $\frac{n^r-nr+r-1}{n-1}$?My try:$(n^r-1)!$=$(n^r-1)(n^r-2)(n^r-3)\cdot\cdot\cdot\cdot3.2.1$.But what next can i do?
What is the highest power of n in $(n^r-1)!$ where $n$, $r$ are positive integers? The answer supplied is $\frac{n^r-nr+r-1}{n-1}$   The highest power of $n$ in $n^r!  = r +$ the highest power of $n$ in $(n^r-1)!$ as $n^r$ contributes $r$ powers.    So,  the highest power of $n$ in $n^r!$ needs to be $\frac{n^r-nr+r-1}{n-1} + r =\sum_{1≤t&lt;r}n^t $.    This is clearly true if $n$ were prime according to .  Again,  if $r=1$, $\frac{n^r-nr+r-1}{n-1}$ becomes $0$.    But according to , the  highest power of composite $n$ in $(n-1)! ≥ 1 $.  So, the answer does not seem to hold for $r=1$ if $n$ is composite. If $n=\prod p_i^{a_i}$, where $p_i$ are distinct primes,  $n^r=\prod p_i^{ra_i}$  So, the highest power $(q_i)$ of $p_i$ in  $n^r!$ is  $$\sum_{1≤t&lt;∞}\left\lfloor\frac{n^r}{p_i^t} \right\rfloor =\sum_{1≤t≤ra_i}\frac{n^r}{p_i^t} +\sum_{ra_i+1≤t&lt;∞}\left\lfloor\frac{n^r}{p_i^t} \right\rfloor$$  as $p^t|n^r$ for $t≤ra_i$.  So, the highest power of $p_i^{a_i}$ in  $n^r!$ is $\left\lfloor \frac{q_i}{a_i} \right\rfloor=P_i$(say).   So, the highest power of n=$\prod p_i^{a_i}$ in  $n^r!$ is $min(P_i)$  I could not proceed anymore.
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