The Order of S_n is n!

See also Symmetric Group & Permutations.

Proof

To show Order (Groups), do an induction. Clearly the base case holds, so consider the inductive step where $|S_n|=n!$. For $S_{n+1}$ we can consider the number of functions that $1$ can be sent to (ie: all $\sigma (1)$) and then consider the $n$-sizes remaining function. There's $n+1$ bijections that $1$ can be sent to, so we multiply this by the $n!$ ones from $S_n$. Thus there are $(n+1)!$ bijections in total, completing the induction.

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