Prove that a cycle of length $k\geq 2$ can be written as a product of $k-1$ transpositions.

Question

Prove that a cycle of length $k\geq 2$ can be written as a product of $k-1$ transpositions as follows: $$ (a_1 ... a_{k-1} a_{k})=(a_1 a_{k})(a_1 a_{k-1})...(a_1 a_2).$$ I found an answer here: Permutations as a product of transpositions but I'm not able to generalize.
It has kind of been an intuition that it will be correct, but I'm not able to present logical mathematical arguements.

Answer 1

Prove by induction. For $k = 2$, $(a_1a_2)$ is already a transposition. Now suppose: $$ (a_1\cdots a_{k-1}a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2) $$ Then: $$ (a_1\cdots a_ka_{k+1}) =^! (a_1a_{k+1})(a_1\cdots a_{k-1}a_k) = (a_1a_{k+1})(a_1a_k)(a_1a_{k-1})\cdots(a_1a_2) $$ So the only non-trivial part of this proof is the equality $=^!$.

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To show this, we want to show that applying the permutation $(a_1\cdots a_ka_{k+1})$ to any $a_i$ is the same as applying the permutation $(a_1a_{k+1})(a_1\cdots a_{k-1}a_k)$. We consider a few cases.

Case 1: $i = k$. Then applying $(a_1\cdots a_ka_{k+1})$ to $a_k$ clearly gives $a_{k+1}$. On the other hand, applying $(a_1\cdots a_{k-1}a_k)$ to $a_k$ gives $a_1$, and composing it with $(a_1a_{k+1})$ gives $a_{k+1}$.

Case 2: $i = k+1$. Applying $(a_1\cdots a_ka_{k+1})$ to $a_{k+1}$ clearly gives $a_1$. $(a_1\cdots a_{k-1}a_k)$ has no effect on $a_{k+1}$, but $(a_1a_{k+1})$ swaps it to give $a_1$.

Case 3: $i = 1,2,\dots,k-1$. Applying $(a_1\cdots a_ka_{k+1})$ to $a_i$ clearly gives $a_{i+1}$, and same for applying $(a_1\cdots a_{k-1}a_k)$. Since $i+1 \neq 1,k+1$, $(a_1a_{k+1})$ has no effect on $a_{i+1}$, so it still gives $a_{i+1}$.

In all cases, both permutations permutate $a_i$ to the same position.