Which is greater, $300 !$ or $(300^{300})^\frac {1}{2}$?

Which is greater among $300 !$ and $\sqrt {300^{300}}$ ?

The answer is $300 !$ (my textbook's answer). I do not know how to solve problems involving such large numbers.

HINT: $(300^{300})^{1/2}=300^{150}$, so you’re comparing

$$\underbrace{300\cdot300\cdot\ldots\cdot300}_{150\text{ factors}}$$

with

$$300\cdot299\cdot\ldots\cdot1=\underbrace{(300\cdot1)\cdot(299\cdot2)\cdot\ldots\cdot(151\cdot150)}_{150\text{ factors}}\;.\tag{1}$$

Show that each of the parenthesized factors in $(1)$ is at least $300$.

When calculations involve large factorials, the use of Stirling approximation for $n!$ is very convenient. In your case, you compare $n!$ to $n^{n/2}$. We can take logarithms of both sides and use the fact that Stirling approximation gives $\log(n!)$ close to $n \log(n) - n$; the logarithm of the second term is simply $\dfrac{n \log(n)} {2}$. Using this approximation, you find that the factorial is the largest as soon as $n > 8$. Without any approximation, the factorial is the largest as soon as $n > 2$ [$n=2$ is the only solution of $n! = n^{n/2}$]

$300!=(300*1) * (299*2) * ... * (151*150)$, 150 pair products total

$(300^{300})^{1/2}= 300^{150} = 300 * 300 * ... * 300$, 150 counts of 300 total

none of pair product in the first line is smaller than $300$, so $300!$ is larger.