Group Theory: Proof of the cycle decomposition of the conjugate The following theorem is proven on this site:

Let $\pi$ and $\rho$ be permutations of $\{ 1,...n\}$. The cycle decomposition of $\rho \pi \rho^{-1}$ is obtained by replacing each integer $j$ in the cycle decomposition of $\pi$ with the integer $\rho (i)$.

I do not understand why $\rho(i)$ lying on the right of $\rho(\pi(i))$ implies that one can just replace every element $i$ in a cycle by $\rho(i)$ and then obtain a cycle decomposition of $\rho \pi \rho^{-1}$. 
Why is that so?
 A: Every permutation can be written as a product of disjoint cycle. So it is enough to prove for cycles. So $\rho$ is a permutation, and we let $\sigma = (i_{1}, i_{2},... i_{r})$. You want to calculate $\rho \sigma \rho^{-1}=\rho (i_{1}...  i_{n})\rho^{-1}$
Now observe if $a \neq \rho(i_{j}) (1\leq j\leq n)$ then $\rho^{-1}(a) \neq i_{j} (1\leq j\leq n)$ , hence $\sigma(\rho^{-1}(a))=\rho^{-1}(a) \implies \rho \sigma \rho^{-1}(a)=a $
And consider $\rho(i_{k})(1\leq k <n)$. $\sigma(\rho^{-1}(\rho(i_{k})))=\sigma(i_{k})=i_{k+1} \implies \rho \sigma \rho^{-1}(\rho(i_{k}))=\rho(i_{k+1})$. And you can easily see $\rho \sigma \rho^{-1}(i_{n})=\rho(i_{1})$.
Hence $\rho (i_{1}...  i_{n})\rho^{-1}= (\rho(i_{1}),\rho(i_{2}),...... \rho(i_{n}))$.
A: You only have to check it for cycles. Write $\pi = (\rho^{-1}(a_1),\rho^{-1}(a_2),\dots, \rho^{-1}(a_n))$. This is possible, because $\rho$ is a permutation. Now let $\rho\pi\rho^{-1}$ act on $a_i$ and you will see that indeed it is mapped to $a_{i+1}$. And that's exactly what you want.  
