What is wrong with this fake proof that $\lim\limits_{n\rightarrow \infty}\sqrt[n]{n!} = 1$?

Question

$$\lim_{n\rightarrow \infty}\sqrt[n]{n!}=\lim_{n\rightarrow \infty}\sqrt[n]{1}*\sqrt[n]{2}\cdots\cdot\sqrt[n]{n}=1\cdot1\cdot\ldots\cdot1=1$$ I already know that this is incorrect but I am wondering why. It probably has something to do with the fact that multiplication in $n!$ is done infinite number of times.

Answer 1

Start by figuring out a simpler example: $$1 = \lim_{n\to\infty} \frac n n = \lim_{n\to\infty} \frac {1+1+\ldots+1} n = \lim_{n\to\infty} \frac 1 n + \frac 1 n + \ldots + \frac 1 n = 0 + 0 + \ldots + 0 = 0$$

Indeed, you cannot exchange sum (or product) and limit if the amount of terms in the sum or product depend on the limiting variable.

Answer 2

Another way of explaining this is that for infinite $n$, each of the factors $\sqrt[n]{1}$, $\sqrt[n]{2}$, $\sqrt[n]{3}$, ... $\sqrt[n]{n}$ will be infinitely close to $1$, but this is not enough to conclude anything about the product because there are infinitely many factors in the product.