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

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?

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