I'm studying and I can't solve this question.
Let be $p, t, n \in \mathbb{Z}^+, p$ prime. Show that $\exists \sigma \in S_n$ of order $p^t \Leftrightarrow p^t \leq n$.
The only thing that I conclude is that if $\sigma$ has order $p^t$ then $p^t$ divides $n!$.
If $p^t\leqslant n$, just take the cycle$$1\mapsto2\mapsto3\mapsto\cdots\mapsto p^t\mapsto1.$$Suppose now that theres's an element $\sigma\in S_n$ whose order is $p^t$. Then $\sigma$ can be expressed as the composition of disjoint cycles $\sigma_1\circ \sigma_2\circ\cdots\circ\sigma_k$. The order of $\sigma$ is the least common multiple of the orders of the $\sigma_j$'s. But we are talking of the power $p^t$ of a prime number here: it can be the least common multiple of a finite set of numbers if and only if one of those numbers is $p^t$. But no cycle in $S_n$ has order greater than $n$. So, $p^t\leqslant n$.