Conjugation of permutation group $S_n$

I want to find the conjugacy classes of the permutation group $S_n$

To start with I think I have to prove that $\pi(\sigma_1\dots \sigma_m)\pi^{-1} = (\pi(\sigma_1)\dots \pi(\sigma_m))$. Where $\pi$ is an arbitrary element and $\sigma$ is a simple cycle.

I have a proof where I basically write out the lhs. I won't post it here as I understand "check my work" question are frowned upon. I was just wondering if there is a more elegant (shorter) proof.

The easiest way to see this is to ask yourself: what happens to $\pi(\sigma_{1})$ when we act on it by $\pi (\sigma_{1} \ldots \sigma_{n})\pi^{-1}$? Well the $\pi^{-1}$ sends us to $\sigma_{1}$, the cycle sends us to $\sigma_{2}$ and then the $\pi$ sends us to $\pi(\sigma_{2})$. This is the definition of the RHS.
This implies that conjugation preserves cycle type in $S_{n}$, so you can find the conjugacy classes by writing down all the permutations with the same cycle type.