Does $p^n$ divide $(p^n-1)!$?

Question

Wilson's Theorem implies that if $p$ is a prime, then $p$ does not divide $(p-1)!$. Can we say something whether $p^n$ divides or not $(p^n-1)!$, where $n$ is any integer at least 3? What about $n=2$?

Answer 1

$(p^n - 1)! = 1*2*3......*(p^n-1)$ so

Note the $p, p^2, p^3,.....,p^{n-1}$ are all less than $p^n-1$.

Also $p, 2p, 3p, .....,p^n-p$ are also less than $p^n-1$.

So $p*2p*3p*...p^2...etc$ will divide $(p^n-1)!$.

In particular $p*p^2*p^3*....*p^{n-1} = p^{1+2+...(n-1)} = p^{\frac {(n-1)n}2}$ will divide $(p^n-1)!$.

(Note: If $p> 2$ then this will not be the largest power of $p$ to divide $(p^n-1)!$ because we are not taking $2p*3p*...*(p-1)p*(p+1)p...$ etc into account. We'll deal with those later.)

So this will be true whenever $\frac {n-1}2n \ge n$ or if $n\ge 3$.

Okay, be what if $n = 2$ or $n=1$? Well obviously if $n =1$ we $(p-1)! = 1*2*....*(p-1)$ and $p$ doesn't divide any $k < p$ so the statement is false for $n=1$.

So what about $n =2$? Well now is when we'll take the $p, 2p, 3p...(p-1)p$ into account.

$(p^2-1)! = 1*2*3....*(p^2 - 1)$ and so $p, 2p,...., (p-1)p$ are all less than $p^2 -1$ so $p*2p*....*(p-1)p = (p-1)!p^{p-1}$ will divide $(p^2 -1)!$.

So if $p-1 \ge 2$ or in other words if $p \ge 3$ then the will be true for $n=2$.

And if $p =2$ and $n=2$? Well, then it is false. $(2^2-1)!= 1*\color{blue}2*(2^2-1)$ and that's not enough copies of $\color{blue}2$.

So conclusion:

False for $n =1$. False for $n = 2;p=2$. And true for all other cases.

Answer 2

It is a well-known and easy-to-prove fact that the multiplicity with which a prime $p$ divides $n!$ is equal to $$\sum_{k=1}^\infty \bigg\lfloor \frac{n}{p^k}\bigg\rfloor$$ Thus, the multiplicity with which $p$ divides $(p^n-1)!$ is equal to $$\sum_{k=1}^\infty \bigg\lfloor \frac{p^n-1}{p^k}\bigg\rfloor=\sum_{k=1}^n (p^{n-k}-1)=\frac{p^n-1}{p-1}-n$$ Which is greater than $n$, for reasonably large $n$ (since it grows exponentially). Thus, for "sufficiently large" $n$, we have that $p^n|(p^n-1)!$.