Limit of n!^{1/n} two ways that give distinct results. Contradiction? I tried two methods to compute the limit, $\lim \limits_{n \rightarrow \infty} \sqrt[n]{n!}$ and i got distinct results. 
Method 1: Observe that $(2n)! \ge \prod \limits_{k=n}^{2n} k \ge n^{n+1}$. Thus, $\sqrt[2n]{(2n)!} \ge \sqrt[2n]{n^{n+1}} \ge \sqrt{n}$. Same case for $k=2n+1$.So $\lim \sqrt[n]{n!} = +\infty$. 
Method 2: We use algebra of limits, giving
$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \ldots (\lim \sqrt[n]{1}) = 1$. Since $\lim \sqrt[n]{n} = 1$. 
What mistake did I make in these proofs? Thank you!
 A: You can't break apart the product like this$$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \cdots (\lim \sqrt[n]{1})$$simply because the number of terms also goes toward infinity. To illustrate this, take the related example $\lim_{n \to \infty}\sqrt[n]{n^n}$, or written out:
$$
1,\quad \sqrt2\cdot \sqrt2,\quad \sqrt[3]3\cdot \sqrt[3]3 \cdot \sqrt[3]3,\quad \sqrt[4]4\cdot \sqrt[4]4\cdot \sqrt[4]4\cdot \sqrt[4]4,\ldots
$$
which can be described as
$$
\lim_{n \to \infty}\prod_{i = 1}^n a_{i,n} = \lim_{n \to \infty}a_{1, n}a_{2, n}\cdots a_{n,n}\tag{*}
$$
where $a_{i,n} = \sqrt[n]{n}$. This product clearly has no limit, but using your second technique you still get $1$ as the answer.
Your sequence can also be described using $\text{(*)}$, but with $a_{i,n} = \sqrt[n]{i}$ instead. The argument is equally false.
A: Your first proof is correct. The second one is not, because you cannot use "algebra of limits" as you did: in the product you write the number of terms is not fixed but is becoming larger and larger (is equal to $n$); so you cannot say say that the limit of the product is equal to the product of the limits. The same kind of things happen with infinite series: you cannot always pass to the limit under the $\sum$ sign. 
