Proof: For any integer $n$ with $n \ge 1$, the number of permutations of a set with $n$ elements is $n!$.
Your proof seems correct, but (in my opinion) it's too elaborate. It obscures the combinatorial simplicity of the argument. I might write the following:
Clearly, there is only one (that is, $1!$) permutation on a set containing a single element.
Suppose now (for the purpose of induction) that there are $m!$ permutations of a set containing $m$ elements. We seek to show that there are $(m+1)!$ permutations of a set containing $m+1$ elements.
We can construct all permutations of the latter set as follows:
- Choose a permutation of the first $m$ elements.
- Insert the final element into the permutation.
By the inductive hypothesis, Step 1 can be completed in $m!$ ways. Step 2 can be completed in $m+1$ ways, since there are $m+1$ locations into which the final element may be inserted. Therefore, we have (by the multiplication principle) that the number of permutations of an $m+1$ element set is equal to $m! \cdot (m+1)$, which is the desired $(m+1)!$.
When we are given $m+1$ items, we separate a special item from the other $m$ items. We sort the $m$ items and fix their order, that give us $m!$ options.
Now we have $m+1$ positions to choose to place our special items and none of them repeats.
Hence we have $m! \cdot (m+1)= (m+1)!$