Sequence help calculus

Use the squeeze theorem to show that n!/(n^n)=0.

1) Note that for n ≥ 2, we have
0 < n!/n^n = (1 * 2 * 3 *... * n)/n^n ≤ (1 * n * n * ... * n)/n^n = 1/n.

Since lim(n→∞) 0 = 0 = lim(n→∞) 1/n, we conclude that lim(n→∞) n!/n^2 = 0.
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2) Note that for n ≥ 3, we have
0 < 2^n/n! = 2^n/(1 * 2 * 3 *... * n) ≤ 2^n/(1 * 2 * 3 *... * 3) = 2^n/(2 * 3^(n-2)) = 3 * (2/3)^(n-1).

Since lim(n→∞) 0 = 0 = lim(n→∞) 3 * (2/3)^(n-1), we conclude that lim(n→∞) 2^n/n! = 0.
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I hope this helps!