Smallest $n$ such that $p^n \nmid (p^2-1)!$

Question

Maybe trivial for number theorists, but not for me: is the title meaningful to ask for ($p$ is a prime)? If so, what's the answer? Thanks

Answer 1

The answer is $n=p$.

To prove it we use the de polignac formula which in this case tells us: $$v_p\left((p^2-1)!\right)=\sum_{i=1}^{\infty}\Bigg\lfloor \frac {p^2-1}{p^i}\Bigg\rfloor=\Bigg\lfloor \frac {p^2-1}{p}\Bigg\rfloor= p-1$$

The claim follows immediately.

Answer 2

As $p^2-1<p^2$, all the factors of $(p^2-1)!$ are smaller than $p^2$, and as $p$ is prime, that means the highest power of $p$ that divides any of the factors is $p$. It's not hard to see how many factors are divisible by $p$ giving the answer to your question.