After playing around with the algebra, I found that the numerator became $\sum_{i}^{n} \log (i)$ while the denominator could be written as $\sum_i^n \log (n)$, but I couldn't get anywhere formally next. Informally, I see how something going from $\frac{2}{3}$ to $\frac{2+5}{3+5} = \frac{7}{8}$ approaches 1, but I still can't get past the fact that $\lim_{x \to \infty} \frac{x!}{x^x} = 0$.
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$$n! \sim \sqrt{2\pi n} \left(\frac n e\right)^n$$
so that
\begin{align*} \frac{\log n!}{\log n^n} \sim \frac{\log \sqrt{2\pi n} + n \log n - n}{n \log n} \to 1 \end{align*}
since the first and last terms are negligible.
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Since $n!\le n$, it is clear that $\log(n!)\le \log(n^n)=n\log n$. We need a bound in the other direction.
We have $\log 2\ge \int_1^2 \log x\,dx$, and $\log 3\ge \int_2^3 \log x\,dx$ and $\log 4\ge \int_3^4\log x\,dx$, and so on. Thus $$\log 2+\log 3+\cdots +\log n\ge \int_1^n \log x\,dx.$$ The integral is not hard to evaluate (by parts). It is $n\log n-n+1$. It follows that $$n\log n-n+1\le \log(n!)\le n\log n.$$ Divide through by $n\log n$, and let $n\to\infty$.
André Nicolas
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