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
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.