Applying Squeeze Theorem to find $\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0$, new proof and question about old proofs.

Question

I've seen several proofs where to find $$\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0$$ They say it is upper bounded by $\frac{1}{n}$, and then apply limits to both sides of the equation. My question is how they determine that upper bound of $\frac 1 n$ just by expanding the sequence and writing that all the terms except $\frac 1 n$ are $\leq 1$? Couldn't someone argue that it could be $\leq (\frac 2 n \cdot \frac 1 n)$? So my question is when do we say that the $\leq 1$ property is actually valid, till what point is it not just being applied to get an easy limit on the right side of the inequality?

Also I wrote an alternative proof for this that doesn't use squeeze theorem instead using the multiplicity of limits. $$\lim_{n \rightarrow \infty} \frac{n!}{n^n} = \lim_{n \rightarrow \infty} \frac{n}{n} \cdot \lim_{n \rightarrow \infty} \frac{n - 1}{n} \ldots \lim_{n \rightarrow \infty} \frac{2}{n} \cdot \lim_{n \rightarrow \infty} \frac{1}{n}$$ Which must be = $0$, since we know for sure that $$\lim_{n \rightarrow \infty} \frac{1}{n} $$ must be zero, rendering any other steps redundant. Is this an acceptable proof?

Answer 1

Your proofs is not right because the number of terms in the r.h.s is also unbounded.

$n!=2\times\cdots\times n\le \underbrace{n\times\cdots\times n}_{n-1\ \text{times}}=n^{n-1}$ Thus $$0\le\frac{n!}{n^n}\le\frac1n$$