If $F(n)$ is the unique number of arranging $n \ge 1$ unique items, prove inductively that $F(n) = n!$

Question

This question doesn't seem so hard. Then again it does. I'm struggling to move past the induction hypothesis.

The base case is obviously $1!$.

Assuming that $F(n) = n!$ , I take the $(n+1)!$ elements. This is where I'm lost because I don't know how to proceed.

Answer 1

Suppose you have to arrange $n+1$ items and you know that you have $n!$ ways to arrange $n$ items.

Pick an item. You have $n+1$ choices for the placement of the item you chose. For each of this you're left to arrange the remaining $n$ items which can be done in $n!$ ways.

Thus the grand total is $n!+\cdots+n!$, $n+1$ times, i.e. $(n+1)n!=(n+1)!$