Order of Product of Disjoint Permutations

Theorem
Let $$S_n$$ denote the symmetric group on $n$ letters.

Let $$\pi$$ be a product of disjoint cycles of orders $$k_1, k_2, \ldots, k_r$$.

Then $$\left|{\pi}\right| = \mathrm{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$$.

Proof
The order of a $k$-cycle is clearly $$k$$.

Now suppose $$\pi = \rho_1 \rho_2 \cdots \rho_r$$ where each $$\rho_s$$ is of order $$k_s$$, and $$\rho_1$$ to $$\rho_r$$ are mutually disjoint.

Let $$t = \mathrm{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$$.

Then $$\pi^t = \rho_1^t \rho_2^t \cdots \rho_r^t$$.

As $$\forall k: 1 \le k \le r: \exists m \in \mathbb{Z}: t = m k$$, we have $$\forall k: \rho_k^t = e$$.

Thus $$\pi^t = e$$ and certainly $$\left|{\pi}\right| \backslash t$$.

But if $$\pi^u = e$$, then each $$\rho_s = e$$ so $$k_s \backslash u$$ and thus $$u \backslash t$$.

Thus $$\left|{\pi}\right| = t = \mathrm{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$$.