... Clearly^2.1

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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