New Concept

Algebra Level 3

Let defined the concept of $n^{!}$ .

Definition of $n^{!}$ .

Let $n \in \mathbb{W}$ , we defined $n^{!}$ using the formula

$n^{!} = n^{n-1^{n-2^{n-3^{....^{3^{2^{1}}}}}}}$

Find $n$ . If $(n-1)^{!} = 1$

3 2 1 4

2 solutions

Paul Ryan Longhas
Mar 26, 2015

$(n-1)^{!} = 1 => n-1 = 1$ or $(n-2)^{!} = 0$

$n-1 = 1 => n = 2$ or $(n-2)^! = 0 => n-2 = 0 => n = 2$

$=> n=2$

Alex Delhumeau
Jun 8, 2015

In this case, the solution is simply the trivial case where $n-1=1\therefore n=2$ . Nice problem, though!