Calculating the limit $\lim((n!)^{1/n})$ Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm, 
$$\lim_{n\to\infty} \ln((n!)^{1/n}) =  \lim_{n\to\infty} (1/n)\ln(n!) =  \lim_{n\to\infty} (\ln(n!)/n)$$
Applying L'hopital's rule: 
$$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))=  \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k))
 = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k)
 = \lim_{n\to\infty} (\ln(n))$$
I proceeded to expand the $\ln(n)$ out into Maclaurin form
$$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$
Since I $\ln$'ed in the beginning, I proceeded to e the infinity
 $$= e^\infty
 = \infty$$
So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it.

Let a $n\in \Bbb N$. By definition $$[\frac n2]\leq \frac n2<[\frac n2]+1.$$
Then $n!=1\cdot 2\cdot ...\cdot[\frac n2]\cdot ([\frac n2]+1)\cdot...\cdot n>(\frac n2)^{n-[\frac n2]+a}>(\frac n2)^{\frac n2 +a}$, so $(n!)^{\frac 1n}>(\frac n2)^{\frac 12 + \frac an}\to \infty$, thus $(n!)^{\frac 1n}\to \infty.$ We set $a:=0$, if $n$ is even, and $a:=1$, if it is odd.
