[proofplan]
We prove the equality by showing that the two permutations act identically on every element of $\{1,\dots,n\}$. Let $\sigma$ denote the cycle $(a_1\ \cdots\ a_k)$ and evaluate the conjugate $\tau\sigma\tau^{-1}$ at an arbitrary element $x$. If $x$ is one of the relabelled entries $\tau(a_i)$, the conjugate sends it to the next relabelled entry; if $x$ is not one of them, the conjugate fixes it. These two cases exactly describe the cycle $(\tau(a_1)\ \cdots\ \tau(a_k))$.
[/proofplan]
[step:Define the cycle and the relabelled cycle as permutations of $\{1,\dots,n\}$]
Let $\sigma \in S_n$ denote the $k$-cycle
\begin{align*}
\sigma=(a_1\ \cdots\ a_k).
\end{align*}
Thus the entries $a_1,\dots,a_k$ are distinct elements of $\{1,\dots,n\}$, and $\sigma$ acts by
\begin{align*}
\sigma(a_i)=a_{i+1}
\end{align*}
for $1 \le i < k$, by
\begin{align*}
\sigma(a_k)=a_1,
\end{align*}
and fixes every element of $\{1,\dots,n\}\setminus\{a_1,\dots,a_k\}$.
Define $\rho \in S_n$ by
\begin{align*}
\rho=(\tau(a_1)\ \cdots\ \tau(a_k)).
\end{align*}
Since $\tau$ is a bijection of $\{1,\dots,n\}$, the elements $\tau(a_1),\dots,\tau(a_k)$ are distinct. Hence $\rho$ is the cycle sending $\tau(a_i)$ to $\tau(a_{i+1})$ for $1 \le i < k$, sending $\tau(a_k)$ to $\tau(a_1)$, and fixing every element outside $\{\tau(a_1),\dots,\tau(a_k)\}$.
[/step]
[step:Compute the conjugate on the relabelled cycle entries]
Let $x \in \{\tau(a_1),\dots,\tau(a_k)\}$. Then there is a unique index $i \in \{1,\dots,k\}$ such that
\begin{align*}
x=\tau(a_i).
\end{align*}
If $1 \le i < k$, then $\tau^{-1}(x)=a_i$, so
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\tau(\sigma(a_i))=\tau(a_{i+1})=\rho(x).
\end{align*}
If $i=k$, then $\tau^{-1}(x)=a_k$, so
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\tau(\sigma(a_k))=\tau(a_1)=\rho(x).
\end{align*}
[guided]
We first check the points that should move under the conjugated cycle. Take an element $x \in \{\tau(a_1),\dots,\tau(a_k)\}$. Because $\tau$ is injective and the entries $a_1,\dots,a_k$ are distinct, there is a unique index $i \in \{1,\dots,k\}$ with
\begin{align*}
x=\tau(a_i).
\end{align*}
Now evaluate the conjugate from right to left. If $1 \le i < k$, then applying $\tau^{-1}$ removes the relabelling:
\begin{align*}
\tau^{-1}(x)=a_i.
\end{align*}
The cycle $\sigma=(a_1\ \cdots\ a_k)$ sends $a_i$ to the next entry $a_{i+1}$, and applying $\tau$ relabels that next entry. Therefore
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\tau(\sigma(a_i))=\tau(a_{i+1}).
\end{align*}
By the definition of $\rho=(\tau(a_1)\ \cdots\ \tau(a_k))$, we also have
\begin{align*}
\rho(x)=\rho(\tau(a_i))=\tau(a_{i+1}).
\end{align*}
Thus $(\tau\sigma\tau^{-1})(x)=\rho(x)$ in this case.
The final entry wraps around. If $i=k$, then $\tau^{-1}(x)=a_k$, and the defining action of $\sigma$ gives $\sigma(a_k)=a_1$. Hence
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\tau(\sigma(a_k))=\tau(a_1).
\end{align*}
The cycle $\rho$ also sends its last listed entry $\tau(a_k)$ to its first listed entry $\tau(a_1)$, so
\begin{align*}
\rho(x)=\rho(\tau(a_k))=\tau(a_1).
\end{align*}
Thus $(\tau\sigma\tau^{-1})(x)=\rho(x)$ for every relabelled cycle entry $x$.
[/guided]
[/step]
[step:Compute the conjugate outside the relabelled cycle entries]
Let $x \in \{1,\dots,n\}\setminus\{\tau(a_1),\dots,\tau(a_k)\}$. Then $\tau^{-1}(x)\notin \{a_1,\dots,a_k\}$; otherwise, if $\tau^{-1}(x)=a_i$ for some $i$, then $x=\tau(a_i)$, contradicting the choice of $x$.
Since $\sigma$ fixes every element outside $\{a_1,\dots,a_k\}$, we have
\begin{align*}
\sigma(\tau^{-1}(x))=\tau^{-1}(x).
\end{align*}
Applying $\tau$ gives
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\tau(\tau^{-1}(x))=x.
\end{align*}
By definition, $\rho$ also fixes every element outside $\{\tau(a_1),\dots,\tau(a_k)\}$, so
\begin{align*}
\rho(x)=x.
\end{align*}
Therefore $(\tau\sigma\tau^{-1})(x)=\rho(x)$ for every such $x$.
[/step]
[step:Conclude equality from equality of values on every element]
The set $\{1,\dots,n\}$ is the disjoint union of $\{\tau(a_1),\dots,\tau(a_k)\}$ and its complement. The previous two steps show that
\begin{align*}
(\tau\sigma\tau^{-1})(x)=\rho(x)
\end{align*}
for every $x \in \{1,\dots,n\}$. Hence the two permutations are equal as elements of $S_n$:
\begin{align*}
\tau\sigma\tau^{-1}=\rho.
\end{align*}
Substituting $\sigma=(a_1\ \cdots\ a_k)$ and $\rho=(\tau(a_1)\ \cdots\ \tau(a_k))$ gives
\begin{align*}
\tau(a_1\ \cdots\ a_k)\tau^{-1}=(\tau(a_1)\ \cdots\ \tau(a_k)).
\end{align*}
[/step]
