Given $\tau\in{S_n} $ a cycle of length $m$ that $\tau^{m} = e$

Question

I'm trying to rigorously prove that given a cycle $\tau\in{S_n} $ of length $m$, then $\tau^{m} = e$ where $e$ is identity.

The funny thing is that I know why it works and understand it intuitively but when I try to prove it using the tools I have I cannot come up with something solid enough.

What I've tried so far:

1. Tried induction : Starting with k=2 (transposition) it easy enough to show that it works then assuming that $\tau$ is of length $m$ that $\tau^{m} = e$ tried to prove that given $\tau_2$ with length $m+1$ that $\tau_2^{m+1} = e$ but didn't come up with a way to use my "induction hypothesis".
2. We did prove that every cycle of length $m$ can be represented as $m-1$ transpositions, tried to express $\tau=\tau_1\circ \tau_2\circ \cdots \circ \tau_{k-1}$ where $\tau_1,\cdots,\tau_{k-1}$ are transpositions but no luck there too.

Answer 1

I think the best way to prove this would be to look at a permutation $\tau \in S_n$ as a function $\tau:\mathbb{Z}_n\to\mathbb{Z}_n$, and then a cycle of length $m$ is just a function that satisfies $\tau(a_i) = a_{(i+1 \mod m)}$ for some $\{a_0,\ldots,a_{m-1}\}$, and doesn't change any $a\notin \{a_0,\ldots,a_{m-1}\}$.

Then, $\tau^m$ is simply applying $\tau$ $m$ times, which means that $$\tau^m(a_i) = \tau^{m-1}(a_{(i+1\mod m)}) = \tau^{m-2}(a_{(i+2\mod m)}) = \ldots = a_{(i+m\mod m)} = a_i$$ for all $\{a_0,\ldots,a_{m-1}\}$, and clearly for any $a\notin \{a_0,\ldots,a_{m-1}\}$, $\tau^m(a)=a$, meaning that $\tau^m = e$.

Answer 2

Let $\tau=(a_0\ a_1\ \cdots\ a_{m-1})$. Then for each $i$ we have $\tau(a_i)=a_{i+1}$, where the indices are taken modulo $m$ (for example $a_m=a_0$). Hence $\tau^2(a_i)=\tau\bigl(\tau(a_i)\bigr)=\tau(a_{i+1})=a_{i+2}$. Continuing in this way we get $\tau^m(a_i)=a_{i+m}=a_i$, and so $\tau^m$ fixes the numbers $a_i$. This concludes the proof because obviously $\tau$ and all of its powers fix the numbers in $\{1,\dots,n\}\setminus\{a_1,\dots,a_m\}$.