$\frac{N\log{N}}{k\log{k}}\approx  \log_{k!}{N!}$  What is the simple way to show that $$\frac{N\log{N}}{k\log{k}}\approx  \log_{k!}{N!}\quad?$$
I tried to use the factorial and the log rules but.. 
Thanks.
 A: The identity isn't true. Assuming the logs are natural, take $k = e$ and $N = 2$. We have 
\begin{align}
0.509989... \approx \tfrac{2}{e} \log 2 = \frac{2 \log 2}{e \log e} \neq \log_{e!} 2 = \tfrac{\log 2}{\log e!} \approx 0.47281...
\end{align}
Edit: I see that you replaced an equality with an approximation. In this case, it depends on the relative size of $N$ compared to $k$. To see this, just use Stirling's Approximation (on both of the factorials)
\begin{align}
N! \sim \sqrt{2 \pi N} \left( \frac{N}{e} \right)^{N}.
\end{align}
That is,
\begin{align}
\frac{\log N!}{\log k!} \approx \frac{N \log N + \tfrac{1}{2} (\log 2 \pi N) - N}{k \log k + \tfrac{1}{2} (\log 2 \pi k) - k}
\end{align}
Can you finish the argument? What else do you know about $N$ and $k$?
A: Noting that $\log _{k!} N! = \frac{{\log N!}}{{\log k!}}$, we want to show
$$
\frac{{N\log N}}{{k\log k}} \approx \frac{{\log N!}}{{\log k!}}.
$$
Let us interpret this approximation as
$$
\frac{{\frac{{\log N!}}{{\log k!}}}}{{\frac{{N\log N}}{{k\log k}}}} \approx 1,
$$
for sufficiently large $n$ and $k$. From the simple inequality*
$$
n\log n - n + 1 \le \log n! \le (n + 1)\log (n + 1) - n,
$$
we get
$$
\frac{{N\log N - N + 1}}{{(k + 1)\log (k + 1) - k}} \le \frac{{\log N!}}{{\log k!}} \le \frac{{(N + 1)\log (N + 1) - N}}{{k\log k - k + 1}}.
$$
Hence
$$
\frac{{\frac{{N\log N - N + 1}}{{(k + 1)\log (k + 1) - k}}}}{{\frac{{N\log N}}{{k\log k}}}} \le \frac{{\frac{{\log N!}}{{\log k!}}}}{{\frac{{N\log N}}{{k\log k}}}} \le \frac{{\frac{{(N + 1)\log (N + 1) - N}}{{k\log k - k + 1}}}}{{\frac{{N\log N}}{{k\log k}}}},
$$
or
$$
\frac{{N\log N - N + 1}}{{N\log N}}\frac{{k\log k}}{{(k + 1)\log (k + 1) - k}} \le \frac{{\frac{{\log N!}}{{\log k!}}}}{{\frac{{N\log N}}{{k\log k}}}} \le \frac{{(N + 1)\log (N + 1) - N}}{{N\log N}}\frac{{k\log k}}{{k\log k - k + 1}}.
$$
From this follows clearly the desired approximation.


*

*Obtained by bounding the sum $\sum\nolimits_{x = 1}^n {\log x} = \log n!$ with an integral from above and below; see Wikipedia for more details.

