If $h$ is the $k$-cycle and g element of $S_n$, prove that $ghg^{-1}=(g(a_1)g(a_2)...g(a_k))$

Question

$g$ element of $S_n$, prove that $g(a_1 a_2 ... a_k)g^{-1}=(g(a_1)g(a_2)...g(a_k))$

I'm more concerned with how I can derive the prove of this question

Answer 1

Hint:

$$g(a_1a_2\dots a_k)g^{-1}(g(a_i))=g(a_1a_2\dots a_k)(a_i)=g(a_{i+1})$$