You can try the definition of convergence.
We want to show that n!/n^n converges to 0.
So we’d like to show that for any epsilon > 0,
we can find some positive integer M such that
for all positive integers N > M, we’d have that
N!/N^N is less than epsilon.
This would show that n!/n^n converges to 0.
Proof:
Let epsilon be some arbitrarily small positive number less than 1/2.
We know that if we wanted to, we could find some positive integer J
such that 0 < 1/J < epsilon, since the sequence {1/n} converges to 0.
We wanted to show that we can find some positive integer M such that
for all positive integers N > M, we’d have that
N!/N^N is less than epsilon.
Instead, we will show that we can find some positive integer M such that
for all positive integers N > M, we’d have that
N!/N^N is less than 1/J.
This would imply that N!/N^N < epsilon.
My claim is that you can let M be J, because
J!/J^J is less than 1/J for all J > 2 (which holds since epsilon is less than 1/2).
We can show that J!/J^J < 1/J by showing that J! < J^(J-1),
which we can do by induction (omitted).
Therefore, for any epsilon > 0,
we can find some positive integer M such that
for all positive integers N > M, we’d have that
N!/N^N is less than epsilon.
So n!/n^n converges to 0.