Which of the numbers $300!$ and $100^{300}$ is greater Determine which of the two numbers $300!$ and $100^{300}$ is greater.
My attempt:Since numbers starting from $100$ to $300$ are all greater than $100$. But am not able to justify for numbers between $1$ to $100$
 A: $n! > (n/e)^n$
so
$300!
>(300/e)^{300}
> 100^{300}
$
since $e < 3$.
To show
$n! > (n/e)^n$,
it is true for
$n=1, 2$, and $3$.
If it is true for $n$
and false for $n+1$,
then
$n! > (n/e)^n$
and
$(n+1)! \le ((n+1)/e)^{n+1}$
so,
dividing,
$n+1
< \frac{((n+1)/e)^{n+1}}{(n/e)^n}
= \frac{(n+1)^{n+1}}{en^n}
$
or
$e < (1+1/n)^n
$
which is false.
Note that this proof 
can be easily modified
to use
$e > (1+1/n)^n
$
to show that
$n! > (n/e)^n$
implies that
$(n+1)! > ((n+1)/e)^{n+1}$.
A: Using Ross Millikan's suggestion and Ivoirians's idea, let us consider $$f(n)=\log_{100}(n!)-n$$ Now, let us use Stirling approximation for $n!$; this gives $$f(n) =-n+\frac{n (\log (n)-1)}{\log (100)}+\frac{\log (2 \pi n)}{2 \log (100)}+O\left(\sqrt{\frac{1}{n}}\right)$$ So, $$f'(n)\approx \frac{\log (n)}{\log (100)}+\frac{1}{n \log (10000)}-1$$ $$f''(n)\approx \frac{1}{n \log (100)}-\frac{1}{n^2 \log (10000)}$$ The second derivative is positive for any value of $n>1$.
The first derivative cancels at $$n_*=100 e^{W\left(-\frac{1}{200}\right)}$$ where appears Lambert function which can be approximated again; so $n_*\approx 99.5$ which corresponds to a minimum of $f(n)$. Now, a look at the function $f(n)$ shows that it is negative for $n<268$ and positive for any larger value of the argument.
Edit
May be, you could be interested by this question of mine which, adapted to your problem, shows that an upper bound ot the solution of $n!=a^n$ is given by $$n=-\frac{\log (2 \pi )}{2 W\left(-\frac{\log (2 \pi )}{2 e a}\right)}$$ which, for large values of $a$, can be approximated by $n\approx e a-\frac{1}{2} \log (2 \pi ) $. For $a=100$, this leads to $n \approx 271$.
A: Take the logarithm of each side, base $100$. Since the $\log$ function is monotonically increasing, this preserves the ordering. $\log_{100}(100^{300})$ is clearly $300$. Meanwhile,
$$\log_{100}300! = \log_{100}1 + \log_{100}2 + \dotsb + \log_{100}300 = \sum_{n=1}^{300} \log_{100}n .$$
We can try using integrals to approximate this series. Since, again, $\log$ is monotonically increasing, we have:
$$\sum_{n=1}^{300} \log_{100}n \geq \int_{1}^{300} \log_{100}x\, dx = \dfrac{1}{\ln 100} \int_{1}^{300} \ln x\, dx = \dfrac{1}{\ln 100} \left[x \ln x - x \right]^{300}_1 = \dfrac{300\ln 300 - 299}{\ln 100} = 306.64... > 300.$$
This implies that $300! \geq 100^{300}.$
A: Consider $$k! > n^k$$ and note that $$\lim_{k\to \infty} \frac{(k!)^{1/k}}{k} = \frac{1}{e}$$ 
This implies that if $n$ is smaller than $\frac{k}{e}$, the first statement is true. So if we take $n = 100$ and $k = 300$, we get
$$100 < \frac{300}{e}$$
and $$300! > 100^{300}$$
A: $\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\cdots $
$$e^x>\frac{x^n}{n!}$$  for $x=n$
$$n!>\bigg(\frac{n}{e}\bigg)^n>\bigg(\frac{n}{3}\bigg)^n$$
for $n=300.$  we have $\color{red}{300!>(100)^{300}}$
