We investigate on an arbitrary $a_i$ : $c(a_1 \ a_2 \dots \ a_k)c^{-1}(a_i)$.
First step, $c(a_i)=a_k$. Second step, $(a_1 \ a_2 \dots \ a_k)(a_k)=a_{k+1}$, Third step, $c^{−1}(a_{k+1})=? $.
Any answer that I read in MSE was not helpful to understand. In the final step, they all imply $c^{−1}(a_{k+1})=c(a_i)$, but why $c^{−1}(a_{k+1})=a_k=(a_1 \ a_2 \dots \ a_k)^{−1}(a_{k+1})$? which may imply $c=(a_1 \ a_2 \dots \ a_k)$ !
I would appreciate any simple clear detailed explanation.
Your question is about permutations and how conjugating a $k$-cycle by a permutation preserves the cycle type.
Claim: Let $c\in S_n$ and let $(a_1 \; a_2 \; \ldots \; a_k)$ be a $k$-cycle in $S_n$. Then $$ c (a_1 \; a_2 \; \ldots \; a_k) c^{-1} = (c(a_1) \; c(a_2) \; \ldots \; c(a_k))$$
Proof: Let's just consider how the left and right hand sides act on $c(a_1)$.
Then right hand side sends $c(a_1)$ to $c(a_2)$.
$c^{-1}$ sends $c(a_1)$ to $a_1$, which the $k$-cycle sends to $a_2$, which $c$ then sends to $c(a_2)$. Thus the left hand side as the composition of those three operations sends $c(a_1)$ to $c(a_2)$.
We chose $c(a_1)$ without loss of generality so we obtain the same agreement if we pick any $c(a_i)$. If a number in the set $\{ 1,\dots , n \}$ is not in the $k$-cycle, so it is not one of the $a_i$, then it is clearly fixed by both sides. So both sides are the same.