Originally Posted by

**SlipEternal** Consider the set $\displaystyle \{1,2,\ldots,p^n\}$. Every p integers is divisible by p. There are $\displaystyle \dfrac{p^n}{p} = p^{n-1}$ such numbers. Every integer divisible by $\displaystyle p^2$ is also divisible by $\displaystyle p$, so you need to cound those numbers an additional time (since the factorial takes a product of all of those numbers). There are $\displaystyle \dfrac{p^n}{p^2} = p^{n-2}$ integers with $\displaystyle p^2$ as a factor. And in general, there are $\displaystyle p^{n-m}$ integers with $\displaystyle p^m$ as a factor. So, add that all up:

$\displaystyle \text{ord}_p\left( p^n\right)! = \sum_{m = 1}^n p^{n-m}$

For b) hint: (1) the answers should be 3 and 5. (2) all p

Edit: for b) (1), I think $\displaystyle p=\infty$ works, too