Solve in $\mathbb R$ the factorial

Notice that if $n=0$ we have then $0 =1$ which is absurd and thus $n\not = 0$ . $\,(2n)! = 1 + n! \implies \,2(2n-1)! = 1+n!$ Note that $n!$ is even for $n\geq 2$ which implies that $1+n!$ is an odd every integer $\forall n \geq 2$ . Since the left hand side is even , the value that satisfy the respective sides $n\leq 2 =1$ .