This question has been asked before but I did not understand the accepted answer and it was 8 years old so I did not want to comment on it.
I want to understand what is incorrect in my working of the following-
$$ \lim \limits_{n \to \infty} (n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}[n^{\frac{1}{n}}\cdot (n-1)^{\frac{1}{n}} \cdot (n-2)^{\frac{1}{n}} \cdot \ldots\cdot 1^{\frac{1}{n}}] \tag{1}$$
Now, the product rule of limits says, $$ \lim\limits_{x \to a}[f(x) \cdot g(x)] = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a} g(x)$$
Let $f(n) = n^{\frac{1}{n}}$, and the remaining be $g(n)$. Using this in $(1)$, I get-
$$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}n^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}[(n-1)^{\frac{1}{n}} \cdot (n-2)^{\frac{1}{n}} \cdot \ldots \cdot 1^{\frac{1}{n}}]$$
Repeating the step and continuously breaking the rightmost limit using product law of limits, I get-
$$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}n^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-1)^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-2)^{\frac{1}{n}} \cdot \ldots \cdot \lim\limits_{n \to \infty}1^{\frac{1}{n}}$$
Now, I know the limit of each of these individual terms is $1$. So, I write it as- $$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = 1 \cdot 1 \cdot 1 \cdot \ldots \cdot 1 = 1$$
I know this answer is incorrect as the correct limit is $+\infty$, which can be seen through Stirling's Approximation, but I don't know what is incorrect in my working.
Best Answer
You cannot use the product rule as the number of terms in the sequence is not fixed. Instead, if you consider $a_n$ as the sequence, then as the terms of the sequence are positive you have: $a_n = \text{exp}(\log(a_n)) = \exp(\frac{1}{n}\sum^{n}_{i=2}\log(i))$. Now since $\log$ is strictly increasing we have(when $n$ is even),
$a_n =\exp(\frac{1}{n}\sum^{n}_{i=2}\log(i)) \geq \exp(\frac{1}{n}\sum^{n/2}_{i=2}\log(i)) + \frac{1}{2}\log(\frac{n}{2})) \geq \exp(\frac{1}{n}\sum^{n/2}_{i=2}\log(2)) + \frac{1}{2}\log(\frac{n}{2}))\geq \sqrt{\frac{n}{2}}\exp(1/6), n \geq 4$.
Hence the sequence is unbounded.