Prove that P(n) = n!< (n^n) for n>1

I got stuck...

This is what I have so far

a) show p(2) is true
2!<2^2
2<4 so this part is done.

b) so my inductive hypothesis is as follows
m!1

c) We need to prove p(m+1)
(m+1)!<(m+1)^m+1 , m>1

now I think this is next:
(m+1)! + (m+2)! < (m^m) + (m+2)!

I dont know what to do next....

let me continue from c)

m^m < (m+1)^m [since (m+1)^m = m^m + other positive terms]
since m! so (m+1)*m! < (m+1)(m+1)^m [i just multiplied both sides of the inequality by m+1 which is positive]
so (m+1)! < (m+1)^(m+1) [using basic math]

so the hypothesis holds

take the log of both sides:

log n! = log n + log (n -1) + ... + log 2

log n^n = n * log n, now you can see which is bigger easily. since log n > log n -1 for all n > 1. right?

You shouldn't be adding.

(m+1)! = m!(m+1) = mm! + m! < m(m^m) + (m^m) = (m+1)(m^m) < (m+1)((m+1)^m) = (m+1)^(m+1)