In symmetric group $S_n$, $g$ and $g^k$ are conjugate if $(m,k)=1$ where $m$=order of $g$

Question

Let $S_n$ the symmetric group. I'd like to show that for every $g\in S_n$, if $m$ is the order of $g$

$g^k$ and $g$ are conjugate if $(k,m)=1$

It's well-know that $\tau, \sigma \in S_n$ are conjugate if only if they're are the same cyclic type. So let $g$ be written as products of disjoint cycles as $g:=\alpha_1 \cdots \alpha_l$ where $\alpha_i$ are $k_i$-cycles. Why $$g^k:=(\alpha_1 \cdots \alpha_l)^k=\beta_1 \cdots \beta_l?$$ where $\beta_i$ are $k_i$-cycles.