Why $c_1 ^{\omicron(\sigma)} = c_1 \implies c_1(\tau _1) = \tau_1 $ etc.

So let's say we have a permuation $\sigma$ analysed into disjoint circles $c_1, c_2, .., c_k$, i.e., $\sigma = c_1 c_2..c_k$.

Now let's denote $c_1 = (\tau_1, \tau_2, .., \tau_i)$ and $\omicron(\sigma)$ as the order of $\sigma$.

In a proof, at some point we reached the conclusion that $c_1 ^{\omicron(\sigma)} = c_1$ and my professor immediately wrote that $ \implies c_1^{\omicron(\sigma)}(\tau _1) = \tau_1 , .., c_k^{\omicron(\sigma)}(\tau _k) $.

It's not that obvious to me and I am trying to prove it.

It should be

$$c_j^{\omicron(\sigma)}={\rm id}$$

for all $j$, where ${\rm id}$ is the identity function on the underlying set the permutations act on. But then

$$c_j^{\omicron(\sigma)}(\tau_r)={\rm id}(\tau_r)=\tau_r$$

for all $r$.