How many permutations of $[n]$ are cycles of length $n$?

Question

Note that $[n]=\{1,2,\dots,n\}$. A cycle of length $n$ permutation is when $1, f(1), f^2(1), \dots, f^n(1)$ range over all of $[n]$. How many permutations of $[n]$ are cycles of length $n$?

I was thinking of the circular permutation...

Answer 1

You have $n-1$ options to choose the value of $f(1)$, it can't be $1$. Then you have $n-2$ options to choose $f^2(1)$, because it can't be $1$ and can't be $f(1)$. Then $n-3$ options to choose the value of $f^3(1)$. Continue this way and you will get the number of $n$-cycles is $(n-1)!$.

Answer 2

In cycle notation, ponder the ways to fill all slots, thinking of the cycles as vectors. Clearly there are $n!$ of them. However, any shift-and-wraparound of a vector is a different vector but the same cycle. Since there are $n$ such shifts, the answer is $n!/n= (n-1)!$.