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

$$\lim_{n\rightarrow \infty}\sqrt[n]{n!}=\lim_{n\rightarrow \infty}\sqrt[n]{1}*\sqrt[n]{2}\cdots\cdot\sqrt[n]{n}=1\cdot1\cdot\ldots\cdot1=1$$ 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.

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Somewhat similar to this one. In that question that number of summands changes with $n$, in this question it is number of factors. –  Martin Sleziak May 30 '13 at 12:12
Perhaps it is also appropriate to add at least one link to the correct proof. –  Martin Sleziak May 30 '13 at 13:41
What's wrong is that $\lim_{n\rightarrow \infty}{\sqrt[n]{1} \cdot \sqrt[n]{2} \cdots \sqrt[n]{n} = \lim_{n\rightarrow \infty}{1^n}}$ Which is an indeterminate form. –  Bakuriu May 30 '13 at 14:49
$1^\infty$ is indeterminate. –  ᴊ ᴀ s ᴏ ɴ Jul 2 '13 at 13:07

Start by figuring out a simpler example: $$1 = \lim_{n\to\infty} \frac n n = \lim_{n\to\infty} \frac {1+1+\ldots+1} n = \lim_{n\to\infty} \frac 1 n + \frac 1 n + \ldots + \frac 1 n = 0 + 0 + \ldots + 0 = 0$$
$0\cdot\infty$ is indeterminate. –  ᴊ ᴀ s ᴏ ɴ Jul 2 '13 at 13:06
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.