Let $p ∈ \mathbb Z$ be a prime and let $k ≥ 1$. Describe all of the nilpotent elements in $\mathbb Z/p^k\mathbb Z$. In particular, how many are there?
Here's my argument: any multiple of $p$ works
Let's see why. Suppose $a = rp$. Then, $a^k = r^kp^k = r^k \cdot 0 = 0$. Now, suppose $a$ isn't a multiple of $p$, when our prime factorization of $a$ doesn't include $p$. No matter what exponent you raise $a$ to, we will never have $(p_1 \cdot \cdots \cdot p_r)^d = r^zp^z$ because if we did, the RHS would have a prime factorization that includes $p$ but the LHS doesn't, and so we have two prime factorizations, which is a contradiction.
Is this correct? If so, how do I "count the multiples of $p$"? Is there a formula / algorithm to determine the multiples of $p$? Or can I just say "the multiples of $p$ up to $p^k$ (since we're working with mod $p^k$)"?
Your approach is correct and could not be shortened considerably. Note that every every $p$-th number is divisible by $p$, hence there are $p^{k-1}$-many of those in the set $\{1, \dots, p^k\}$.