nice and easy

I think this is what u intended....
n!<{n+12}^n....

If that is the case
then:
n!=(1*n)(2*(n-1))*.....

(1+n)²/4 ≥ 1*n
(2+n-1)²/4 ≥ 2*(n-1)
.
.
.
continuing and multiplying
(n+12)^2(n/2) ≥ n!
or n! ≤ (n+12)ⁿ

@asish correct solution.
but equality is not valid.
keep trying there's a simpler solution

oh i forgot to mention n>1 asish your solution is perfect

yeah equality not valid bcz nos are not equal... :)

What abt AM-GM on 1st 'n' naturals?

that's right soumik :D