[Pure maths] Limit of sequence

First, it can be seen that

1 / n^n <= n! / n^n

Also,

n! / n^n
= ( n/n )[ (n-1)/n ][ (n-2)/n ]....{ [ n - ( n - 2 ) ] / n }( 1/n )
= ( 1 - 1/n )( 1 - 2/n ) ... ( 1 - n-2/n )( 1/n )
<= 1 / n

Therefore, 1 / n^n <= n! / n^n <= 1 / n

Lim n->infinty 1/n = 0
Lim n->infinty 1 / n^n = 0

By sandwich principle, Lim n->infinty n! / n^n = 0


Check if my solution is correct or not. I'm not so sure about that.