TITLE: Proving the result $294!<100^{300}<295!$
QUESTION [8 upvotes]: Proving $294!<100^{300}<295!$

$\bf{My\; Try::}$ I have used Stirling Approximation $$n!\approx \left(\frac{n}{e}\right)^n\cdot \sqrt{2\pi n}$$
Put $n=294$ and $n=295$, 
$$294!\approx \left(\frac{294}{e}\right)^{294}\cdot \sqrt{2\pi \cdot 294}$$
and $$295!\approx \left(\frac{295}{e}\right)^{295}\cdot \sqrt{2\pi \cdot 295}$$
Now i did not understand hoe can i solve it, Help required, Thanks

REPLY [2 votes]: Wikipedia tells us that
$$
n\ln \left({\frac {n}{e}}\right)+1\leq \ln n!\leq (n+1)\ln \left({\frac {n+1}{e}}\right)+1
$$
Dividing by $\ln 100$ gives
$$
299.005 < \log_{100} 294! < 301.25
\\
300.24 < \log_{100} 295! < 302.48
$$
which is almost what is needed, except for the upper bound for $\log_{100} 294!$.
The actual values are
$$
\log_{100} 294! \approx 299.82
\\
\log_{100} 295! \approx 301.06
$$
and so we do need tight bounds to solve the question.
Using
$$
{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}<n!<{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}e^{\frac {1}{12n}}
$$
and taking logs as before we get
$$
299.82240 < \log_{100} 294! < 299.82247
$$
which does it.