The following is an excerpt from Gallian's Contemporary Abstract Algebra.
First, observe that a cycle of length $n$ has order $n$. (Verify this yourself.) Next, suppose that $\alpha$ and $\beta$ are disjoint cycles of lengths $m$ and $n$, and let $k$ be the least common multiple of $m$ and $n$. It follows from Theorem $4.1$ that both $\alpha^k$ and $\beta^k$ are the identity permutation $\varepsilon$ and, since $\alpha$ and $\beta$ commute, $(\alpha\beta)^k = \alpha^k\beta^k$ is also the identity. Thus, we know by Corollary 2 to Theorem 4.1 ($a^k=e$ implies that $|a|$ divides $k$) that the order of $\alpha\beta$—let us call it $t$—must divide $k$. But then $(\alpha\beta)^t = \alpha^t\beta^t = \varepsilon$, so that $\alpha^t = \beta^{-t}$ . However, it is clear that if $\alpha$ and $\beta$ have no common symbol, the same is true for $\alpha^t$ and $\beta^{-t}$ , since raising a cycle to a power does not introduce new symbols. But, if $\alpha^t$ and $\beta^{-t}$ are equal and have no common symbol, they must both be the identity, because every symbol in $\alpha^t$ is fixed by $\beta^{-t}$ and vice versa (remember that a symbol not appearing in a permutation is fixed by the permutation). It follows, then, that both $m$ and $n$ must divide $t$. This means that $k$, the least common multiple of $m$ and $n$, divides $t$ also. This shows that $k=t$.
Basically, I would like to know how one proves that $k=t$?
First, $k,t\in\mathbb{N}$. Second, $t$ divides $k$ then $k=ut$ for some $u\in \mathbb{N}$. Also $k$ divides $t$, then $t=vk$ for some $v\in\mathbb{N}$. Therefore, the two relations give $k=ut=vuk$ iff $k(uv-1)=0$. But $k\neq 0$, then $uv=1$ in $\mathbb{N}$. The unique possibility is $u=v=1$. Hence, $k=ut=1t=t$.