Construction of Permutations/Proof 2

Proof
The following is an inductive method of creating all the permutations of $n$ objects.

Base Case
There is clearly one way to arrange one object in order.

Inductive Hypothesis
We assume that we have constructed all $n!$ permutations of $n$ objects.

Induction Step
, let a set $S_n$ of $n$ objects be $\left\{{1, 2, \ldots, n}\right\}$.

Take a permutation of $S_n$:
 * $a_1 \, a_2 \, a_3 \, \ldots \, a_n$

Construct the array:


 * $a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 1 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 3 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 5 2, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \left({n + \dfrac 1 2}\right)$

This will contain $n + 1$ objects.

Now rename the elements of each permutation using the numbers $\left\{ {1, 2, \ldots, n}\right\}$ preserving order.

It is clear that all permutations of $n + 1$ objects can be obtained in this manner, and no permutation is obtained more than once.

As there are ${}^n P_n$ permutations on $n$ objects, there are $\left({n + 1}\right) \times {}^n P_n$ permutations on $n + 1$ objects.

Hence by induction, and the recursive definition of the factorial:
 * ${}^n P_n = n!$