Evaluate the limit $\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)$ 
Evaluate the limit: 
  $$
\lim_{n\to\infty}\log_a\left(\frac{4^nn!}{n^n}\right)\\
a>0\\
a \ne 1
$$

I've started with defining another sequence. Let:
$$
y_n = a^{x_n} = \frac{4^nn!}{n^n}
$$
Consider the fraction:
$$
\frac{y_{n+1}}{y_n} = \frac{4^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{4^nn!}\\
= \frac{4n^n}{(n+1)^n}
$$
Consider the limit:
$$
\begin{align}
\lim_{n\to\infty}\frac{y_{n+1}}{y_n} &= \lim_{n\to\infty}\frac{4n^n}{(n+1)^n} \\
&= \lim_{n\to\infty}4\left(\frac{n}{n+1} \right)^n \\ 
&= {4\over e} > 1
\end{align}
$$
So by this $y_n$ is divergent. Which means:
$$
\lim_{n\to\infty}y_n = \infty
$$
Now I'm having difficulties translating it in a backward direction. We have that:
$$
\lim_{n\to\infty}y_n = \lim_{n\to\infty}a^{x_n} = \infty
$$
Or:
$$
\log_a \lim_{n\to\infty}a^{x_n} = \log_a(\infty)
$$
The answer suggests that:
$$
\lim_{n\to\infty}x_n = 
\begin{cases}
+\infty,\ a > 1\\
-\infty,\ 0 < a < 1
\end{cases}
$$
And I don't see where this appears when going backward from $a^{x_n}$ to $x_n$. Could you please explain that to me?

Hint: Note that $$\log_a{x}=\frac{\ln(x)}{\ln(a)}$$
