[proofplan] Write $\sigma$ as a product of its pairwise disjoint cycles $c_1,\dots,c_r$. Because [disjoint cycles commute](/theorems/774), the $k$-th power of $\sigma$ is the product of the $k$-th powers of the individual cycles. A single cycle of length $m_i$ has $k$-th power equal to the identity exactly when $m_i$ divides $k$. Therefore $\sigma^k$ is the identity exactly for the positive integers $k$ divisible by every $m_i$, and the least such positive integer is $\operatorname{lcm}(m_1,\dots,m_r)$. [/proofplan]

[step:Write the permutation as commuting disjoint cycles] Let $e \in S_n$ denote the identity permutation. For each $i \in \{1,\dots,r\}$, write the cycle $c_i$ as \begin{align*} c_i=(a_{i,1}\ a_{i,2}\ \cdots\ a_{i,m_i}), \end{align*} where $a_{i,1},\dots,a_{i,m_i} \in \{1,\dots,n\}$ are distinct. Define the support of $c_i$ by \begin{align*} A_i:=\{a_{i,1},\dots,a_{i,m_i}\}. \end{align*} The cycles are pairwise disjoint, so $A_i \cap A_j=\varnothing$ whenever $i \neq j$. If $i \neq j$, then $c_i$ fixes every element of $A_j$ and $c_j$ fixes every element of $A_i$, while both cycles fix every element outside $A_i \cup A_j$. Hence $c_i c_j$ and $c_j c_i$ agree on every element of $\{1,\dots,n\}$, so \begin{align*} c_i c_j=c_j c_i. \end{align*} Thus the cycles $c_1,\dots,c_r$ commute pairwise. It follows by induction on $k \in \mathbb{N}$ that \begin{align*} \sigma^k=(c_1c_2\cdots c_r)^k=c_1^k c_2^k \cdots c_r^k. \end{align*} [/step]

[step:Characterize when a power of one cycle is the identity]Fix $i \in \{1,\dots,r\}$. For each integer $q$, interpret the subscript $a_{i,q}$ cyclically modulo $m_i$, so that $a_{i,q}=a_{i,s}$ exactly when $q \equiv s \pmod{m_i}$. By the definition of the cycle $c_i$, for every $j \in \{1,\dots,m_i\}$ and every $k \in \mathbb{N}$, \begin{align*} c_i^k(a_{i,j})=a_{i,j+k}. \end{align*} Therefore $c_i^k$ fixes every element of $A_i$ if and only if $j+k \equiv j \pmod{m_i}$ for every $j \in \{1,\dots,m_i\}$, which is equivalent to \begin{align*} m_i \mid k. \end{align*} Since $c_i$ fixes every element outside $A_i$, this proves \begin{align*} c_i^k=e \iff m_i \mid k. \end{align*}[/step]

[guided]We isolate the basic fact about a single cycle. Fix $i \in \{1,\dots,r\}$, and write \begin{align*} c_i=(a_{i,1}\ a_{i,2}\ \cdots\ a_{i,m_i}). \end{align*} The elements $a_{i,1},\dots,a_{i,m_i}$ form the support \begin{align*} A_i:=\{a_{i,1},\dots,a_{i,m_i}\}. \end{align*} On this support, the cycle advances one position: \begin{align*} c_i(a_{i,j})=a_{i,j+1}. \end{align*} Here the subscript is read cyclically modulo $m_i$, meaning $a_{i,m_i+1}=a_{i,1}$ and, more generally, $a_{i,q}=a_{i,s}$ exactly when $q \equiv s \pmod{m_i}$. Applying the same cycle $k$ times advances $k$ positions, so for every $j \in \{1,\dots,m_i\}$, \begin{align*} c_i^k(a_{i,j})=a_{i,j+k}. \end{align*} Thus $c_i^k$ fixes $a_{i,j}$ exactly when $a_{i,j+k}=a_{i,j}$, which by the cyclic indexing condition is equivalent to \begin{align*} j+k \equiv j \pmod{m_i}. \end{align*} Subtracting $j$ from both sides gives \begin{align*} k \equiv 0 \pmod{m_i}. \end{align*} This congruence is exactly the divisibility condition $m_i \mid k$. The cycle $c_i$ already fixes every element outside $A_i$, and so every power $c_i^k$ also fixes every element outside $A_i$. Therefore $c_i^k$ is the identity permutation on all of $\{1,\dots,n\}$ if and only if it fixes the elements of $A_i$, and this happens if and only if \begin{align*} m_i \mid k. \end{align*} Hence \begin{align*} c_i^k=e \iff m_i \mid k. \end{align*}[/guided]

[step:Convert the identity condition for $\sigma^k$ into divisibility conditions] Let $k \in \mathbb{N}$. From the commuting product formula, \begin{align*} \sigma^k=c_1^k c_2^k \cdots c_r^k. \end{align*} Because the supports $A_1,\dots,A_r$ are pairwise disjoint, the factor $c_i^k$ is the only factor that can move elements of $A_i$. Therefore $\sigma^k=e$ if and only if $c_i^k=e$ for every $i \in \{1,\dots,r\}$. Using the single-cycle criterion from the previous step, this is equivalent to \begin{align*} m_i \mid k \quad \text{for every } i \in \{1,\dots,r\}. \end{align*} Thus, for every $k \in \mathbb{N}$, \begin{align*} \sigma^k=e \iff m_i \mid k \text{ for every } i \in \{1,\dots,r\}. \end{align*} [/step]

[step:Take the least positive exponent that gives the identity] By definition, $\operatorname{ord}(\sigma)$ is the least positive integer $k$ such that $\sigma^k=e$. The previous step shows that this set of positive integers is exactly the set of common positive multiples of $m_1,\dots,m_r$. By definition of least common multiple, the least such positive integer is \begin{align*} \operatorname{lcm}(m_1,\dots,m_r). \end{align*} Therefore \begin{align*} \operatorname{ord}(\sigma)=\operatorname{lcm}(m_1,\dots,m_r). \end{align*} This proves the theorem. [/step]