In Alan F. Beardon's "Algebra and geometry" he asks in an exercise to express $(1\ \ldots\ n)$ as a product of two cycles:
Show that $(1\ 2\ 3\ 4)=(1\ 4)(1\ 3)(1\ 2)$. Express $(1\ 2\ 3\ 4\ 5)$ as a product of $2$-cycles. Express $(1\ 2\ \ldots\ n)$ as a product of $2$-cycles.
I assume that if we are working with the set of numbers $\{1,\ldots,n\}$ doing so is not possible. Since all of the numbers of the set will be part of the cycle $(1\ \ldots\ n)$ already. I am new in this subject, so I do not know if I am making a mistake. Thanks.
Because $(1\ n)(1\ n)$ is the identity, the $n$-cycle $(1\ 2\ \ldots\ n)$ is the product of the two cycles $(1\ n)$ and $$(1\ n)(1\ 2\ \ldots\ n)=(1\ 2\ \ldots\ n-1).$$ Induction on $n$ now also shows that $(1\ 2\ \ldots\ n)$ is a product of $2$-cycles.