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 $100$

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