The strategy is to reduce the problem to cycles and then show each cycle decomposes into transpositions. The [Disjoint Cycle Decomposition](/theorems/775) theorem writes any permutation as a product of disjoint cycles, so it suffices to handle a single $k$-cycle. Step 1 gives an explicit formula expressing any $k$-cycle as a product of $k - 1$ adjacent transpositions, verified by tracking each element through the composition. Step 2 combines the two reductions. By the [Disjoint Cycle Decomposition](/theorems/775) theorem, every $\sigma \in S_n$ is a product of disjoint cycles. It therefore suffices to show that each $k$-cycle can be written as a product of transpositions. **Step 1: Decompose a $k$-cycle.** Let $(a_1\; a_2\; \ldots\; a_k)$ be a $k$-cycle. We claim: \begin{align*} (a_1\; a_2\; \ldots\; a_k) = (a_1\; a_2)(a_2\; a_3) \cdots (a_{k-1}\; a_k). \end{align*} To verify, apply the right-hand side to each $a_i$. The rightmost transposition that moves $a_i$ is $(a_i\; a_{i+1})$ (for $1 \leq i \leq k-1$), which sends $a_i \mapsto a_{i+1}$. All transpositions to its left fix $a_{i+1}$ (since $a_{i+1}$ does not appear in them as a first entry — it only appears in $(a_i\; a_{i+1})$ and $(a_{i+1}\; a_{i+2})$, the latter being to the right). For $a_k$: the transposition $(a_{k-1}\; a_k)$ sends $a_k \mapsto a_{k-1}$, then $(a_{k-2}\; a_{k-1})$ sends $a_{k-1} \mapsto a_{k-2}$, and so on, cascading down to $a_1$. This matches the action of the cycle. **Step 2: Conclude.** Since every permutation is a product of disjoint cycles (Step 0), and each cycle is a product of transpositions (Step 1), every $\sigma \in S_n$ with $n \geq 2$ is a product of transpositions.