... Gaglione1
Supported by the NRL. Author's address: Mathematics, Department, U.S. Naval Academy, Annapolis, MD 21402.amg@usna.edu. 
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... Clearly2.1
To determine a permutation $f$ as above, there are $n$ possibilities for $a_1$, $n-1$ possibilities for $a_2$, ..., $1$ possibility for $a_n$. By the multiplicative principle of counting (in any combinatorics book), it follows that the number of possibly permutations $f$ is $n\cdot (n-1)...\cdot 1=n!$. 
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