Let $p$ be a prime. Prove that $Sym(p)$ is generated by any transposition and any p-cycle. Without loss of generality, we may assume the p-cycle to be $(1\space 2\space \dots\space p)$ just by 'relabelling' the entries with a bijection in any p-cycle. Again, without loss of generality, we actually want to prove that $Sym(p)=<(1\space k), (1\space 2\space \dots\space p)>$, where $k$ can take any value from $2$ to $p$, because even if the first entry of the transposition is not '$1$', say, it's '$2$', then since $(1\space 2\space \dots\space p)=(2\space 3\space \dots\space p\space 1)$, we can again 'relabel' the entries with a bijection.
Since we already know that $Sym(p)=<(1\space 2), (1\space 2\space \dots\space p)>$. Why do $(1\space k)$ and $(1\space 2\space \dots\space p)$ generate $(1\space 2)$ for arbitrary $k$?