Which of the numbers $300!$ and $100^{300}$ is greater

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 $300! \geq 100^{300}.$

$n! > (n/e)^n$ 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 $(n+1)! > ((n+1)/e)^{n+1}$.

$\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 $\color{red}{300!>(100)^{300}}$