Factorials Within Factorials

$(n!)!=n!(n-1)!$

$\Rightarrow~ \dfrac{(n!)!}{n!}=(n-1)!$

$\Rightarrow~ (n!-1)!=(n-1)!$

$\Rightarrow~ n!-1=n-1$

$\Rightarrow~n!-n=0$

$\Rightarrow~n((n-1)!-1)=0$

Therefore we have two cases:

Case 1: $n=0$ .

However we are given that $n$ is a positive integer so we reject this.

Case 2: $(n-1)!-1=0 ~\Rightarrow~ (n-1)!=1$ .

A simple check reveals that $(1-1)!=0!=1$ , $(2-1)!=1!=1$ , and any integer above $n=2$ will not satisfy the equation. Thus $n=1$ , $n=2$ are the only possibilities.

Therefore there are $\large\color{#20A900}{\boxed{2}}$ solutions, $n=1$ and $n=2$ .

The equation is equivalent to $\left(n!-1\right)!=\left(n-1\right)!$

Check that $n=1$ and $2$ both work. If $n\geq3$ , then $n!>n\geq3$ , so $n!-1>n-1\geq2$ and thus $\left(n!-1\right)!>\left(n-1\right)!$ , contradiction. Thus, there are only $\boxed{2}$ solutions.

$( 1! )! = 1! \times ( 1 - 1 )! \rightarrow 1! = 1 \times 0! \rightarrow 1 = 1 \times 1 \rightarrow \text { true }$ .

$( 2! )! = 2! \times ( 2 - 1 )! \rightarrow 2! = 2 \times 1! \rightarrow 2 = 2 \times 1 \rightarrow \text { true }$ .

So, $\boxed { 2 }$ positive integers satisfy the equation.