Let $n\in\mathbb{N}$ and consider $\sigma\in S_n$. Then $\exists k\in \mathbb{N}$ such that $\sigma^k = I$. Show this, and also that if $\sigma = \sigma_1...\sigma_m$ where $\sigma_i$'s are disjoint cycles, then the least $k$ such that $\sigma^k = I$ is the LCM of $(l_1,l_2,...,l_m)$ where $l_i$ is the length of the cycle $\sigma_i$.
For the first part, here's what I did:
Let $\sigma=\sigma_1...\sigma_m$ be the disjoint cycle decomposition of $\sigma$ where the length of each cycle $\sigma_i$ is $l_i$. Note that if $\Gamma = (a_1 a_2 a_3...a_l) \in S_{n}$ is a cycle in $S_n$, $l$ is its length. It is easily seen that $\Gamma^l = I$. Now define $t = LCM(l_1,...,l_m)$. Since disjoint cycles commute, we have $$\sigma^t = \sigma_1^t\sigma_2^t....\sigma_m^t$$ As $l_i\vert t$, each $\sigma_i^t = I$ and so $\sigma^t = I$.
To show that $t$ is the least such number, I must show that if $s\in\mathbb{N}$ and $s<t$ then $\sigma^s \neq I$.
Since $t$ is the LCM, we conclude that there exists at least one $i$ such that $l_i$ does not divide $s$, so that $\sigma_i^s \neq I$. I think I should now consider $$\sigma^s = \sigma_1^s\sigma_2^s....\sigma_i^s...\sigma_m^s$$
but I'm stuck here, what do I do next?
P.S.
I have realized that my questions boils down to: If cycles $\sigma_1, \sigma_2,....,\sigma_k$ are disjoint then $\sigma_1\sigma_2...\sigma_k = I$ iff $\sigma_i= I$ for all $1\leq i\leq k$. Any hints?
Although the results we are going to look into can be formulated at a very general level, I shall limit the frame of the discussion to general permutation groups, since you only seem to be familiar with them. It is however very easy to see how the propositions I am about to formulate extend in the general case (of arbitrary groups).
Let us consider an arbitrary set $A$ and the general symmetric group $\Sigma(A)$, structure by which I am referring to the set $\Sigma(A)$ of all permutations of set $A$ with the implicit binary operation of composition of maps. In general, this operation takes the syntax $\sigma \circ \tau$ for two arbitrary permutations $\sigma, \tau \in \Sigma(A)$, however for simiplicity we shall employ the simplified notation $\sigma\tau$ (used for "abstract" multiplications).
To say that $\sigma$ is a permutation of finite order has a very specific meaning in the general sense of group theory, however in our given context we shall be content to understand it as there exists nonzero $n \in \mathbb{N}^{\times}$ such that $\sigma^n=\mathbf{1}_A$. We also introduce the order of the finite order permutation $\sigma$ -- denoted by $\mathrm{ord}(\sigma)$ -- to be the unique natural number $k$ rendering the following relation true: $$\left\{n \in \mathbb{N}^{\times} \mid \sigma^n=\mathbf{1}_A\right\}=k\mathbb{N}^{\times}.$$ Once again, may I point out that this is not the actual definition of the notion of order, but it is an easy to obtain equivalent description of the order of a permutation of finite order (there are also permutations of infinite order; nevertheless, the order of a permutation is always an elements of the set $\mathbb{N}^{\times} \cup \{\aleph_0\}$).
For arbitrary natural numbers $r, s \in \mathbb{N}$ let us denote their least common multiple by $[r; s]$. In case of an arbitrary family $r \in \mathbb{N}^I$ we use the notation $[r_i;]_{i \in I}$ for the least common multiple of the entire family $r$.
With the above mentions in place, let us consider two permutations of finite order $\lambda, \mu \in \Sigma(A)$ which:
Then $\lambda\mu$ is also of finite order and the relation $\mathrm{ord}(\lambda\mu)=\left[\mathrm{ord}(\lambda); \mathrm{ord}(\mu)\right]$ holds.
For simplicity, let us write $m=\mathrm{ord}(\lambda) \neq 0$, $n=\mathrm{ord}(\mu) \neq 0$ and let us introduce $p\colon=[m; n] \neq 0$. This means that $m, n \mid p$ so there exist $k, l \in \mathbb{N}$ such that $p=km=ln$. Since $\lambda$ and $\mu$ commute we have by virtue of a very general rule of algebraic calculus that $(\lambda\mu)^r=\lambda^r\mu^r$ for any $r \in \mathbb{N}$ (even for any $r \in \mathbb{Z}$, but you might not be familiar with the notion of raising to negative powers). This in particular entails that $(\lambda\mu)^{p}=\lambda^{p}\mu^{p}=\left(\lambda^m\right)^k\left(\mu^n\right)^l=\left(\mathbf{1}_A\right)^k\left(\mathbf{1}_A\right)^l=\mathbf{1}_A$. Since $p \neq 0$, this establishes $\lambda\mu$ as a permutation of finite order. Let us set $q\colon=\mathrm{ord}(\lambda\mu)$ and $M=\left\{r \in \mathbb{N}^{\times} \mid (\lambda\mu)^r=\mathbf{1}_A\right\}$.
By virtue of the characterisation of the order (in the finite case), we have $M=q\mathbb{N}^{\times}$ and since -- as we have just seen above -- $p \in M$, we gather that $q \mid p$. We now argue that the converse relation $p \mid q$ also holds in order infer that $p=q$. Since in particular $q \in M$ we have $(\lambda\mu)^q=\lambda^q\mu^q=\mathbf{1}_A$ which entails $\nu\colon=\lambda^q=\left(\mu^q\right)^{-1}=\mu^{-q}$. It is obvious that $\nu \in \left\{\lambda^n\right\}_{n \in \mathbb{Z}} \cap \left\{\mu^n\right\}_{n \in \mathbb{Z}}=\{\mathbf{1}_A\}$, whence $\lambda^q=\mu^{-q}=\mathbf{1}_A$, which is equivalent to $\lambda^q=\mu^q=\mathbf{1}_A$. Since $q \in \mathbb{N}^{\times}$ we have by virtue of the characterisation of the orders $m$ respectively $n$ of $\lambda$ respectively $\mu$ that $m, n \mid q$. By elementary arithmetic (the definition of the least common multiple), this entails $p \mid q$ and establishes our claim.
Before stating the next result, let us define the support of an arbitrary permutation $\sigma \in \Sigma(A)$ as the subset of elements of $A$ it actually permutes: $$\mathrm{Supp}(\sigma)\colon=\{x \in A \mid \sigma(x)\neq x\}.$$ It is easy to prove that two permutations $\lambda, \mu \in \Sigma(A)$ of disjoint supports necessarily commute (I will let you think about it).
The above proposition concerning finite orders generalises to finite families of permutations $\lambda \in \Sigma(A)$ such that:
The way to prove this (somewhat) generalised version is by induction on $|I|$ (the cardinality of $I$). You might not find it straightforward to supply a proof for this on your own, so you can let me know. For the time being I think you have sufficient new material to digest.
The manner in which the generalised version applies to your problem at least I hope will become clear: if you have a finite family $\kappa$ of disjoint cycles then their supports satisfy the condition of pairwise disjointness in the generalised version above and their individual orders are finite, given by their lengths. Thus, their product $\pi$ is also of finite order $r$, number which is equal to the least common multiple of the family of lengths. Bearing in mind the characterisation of the order $r$ of $\pi$, $\pi^n=\mathbf{1}_A$ can only hold for nonzero natural $n$ if $n$ is a multiple of the order $r$ and therefore never in the case $n<r$ (any strictly positive multiple of $r$ being at least equal to $r$).