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I am at a complete loss as to how to even begin. I think it has something to do with the fact that any p-cycle can be represented by $(123...p)$?

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    Just for context, the $n$-cycle $(12...n)$ and a transposition $(ij)$ in $S_n$ generate $S_n$ if and only if $(j-i,n) = 1$. (For instance, $(1234)$ and $(13)$ do not generate $S_4$.) When $n = p$ is prime that gcd condition is automatically satisfied. – KCd May 06 '15 at 05:39
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    Many textbooks explain why $(12)$ and $(123\cdots n)$ generate all of $S_n$. The upshot is that when $n=p$ is a prime, $\alpha=(ab)$ is a transposition, and $\beta$ is a $p$-cycle, then A) all non-identity powers of $\beta$ are themselves $p$-cycles, and B) one of them has the numbers $a$ and $b$ in adjacent positions (in that order) in its cycle presentation. – Jyrki Lahtonen May 06 '15 at 05:40
  • @JyrkiLahtonen So how can I use this to prove it? I understand what you're saying but I don't see yet how it gives me the result. –  May 06 '15 at 06:01
  • @Bob: Locate a proof of that fact. Repeat its argument with roles of the numbers changed: $a$ takes the role of $1$, $b$ the role of $2$, and the rest according to the order the numbers appear in that chosen power of $\beta$. – Jyrki Lahtonen May 06 '15 at 06:12
  • Related: https://math.stackexchange.com/questions/877824 – Watson Dec 25 '16 at 13:39

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