Prove that any conjugate of $a$ has the same order as $a$. [duplicate]

I am trying to prove that any conjugate of $a$ has the same order as $a$.

Let $G$ be a group and let $a \in G$. An element $b \in G$ is called a conjugate of $a$ if $b=xax^{-1}$.

My professor gave us a hint. Prove $a^k=e$ iff $b^k=e$ for all $k \in \mathbb{Z}$ implies $o(a) = o(b)$.

My first move is to $b^k = (xax^{-1})^k = (x^{-1})^ka^kx^k = (x^{-1})^kex^k = (x^{-1})^kx^k=x^{-k+k}=x^0$

I am not sure that I am doing this correctly? I am trying to get $b^k=e$. However when I raise $(xax^{-1})^k$, I am not sure if I what I have done is correct?

I am kind of confused here.

The step $(xax^{-1})^k=x^ka^kx^{-k}$ is wrong. What you should have is $$(xax^{-1})^k=xax^{-1}xax^{-1}\cdots xax^{-1}=xaea\cdots ax^{-1} = xa^kx^{-1}.$$