Which one is greater $600!$ or $300^{600}$ 
Which one is greater $600!$ or $300^{600}$

$\bf{My\; Try::}$ I have used Stirling Approximation.
For large $n>2\;,$ We can write $\displaystyle n! \approx \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$
So $$600!\approx 
\left(\frac{600}{e}\right)^{600}\sqrt{2\cdot \pi \cdot 600}<300^{600}$$
My question is how can we i solve using algebraic Inequalities , Help required
Thanks
 A: Divide them , you will notice $$\frac{600!}{300^{600}}=2\prod_{k=1}^{299}\frac{(300-k)(300+k)}{300^2}=2\prod_{k=1}^{299}\left(1-\frac{k^2}{300^2}\right)$$ Now take $k =299$ out , you get $$\frac{600!}{300^{600}}=2\left(1-\frac{299^2}{300^2}\right)\prod_{k=1}^{298}\left(1-\frac{k^2}{300^2}\right)$$ Now we are multiplying a lot of terms but all of them are less than $1$ so $600!<300^{600}$.
A: Lemma : For $k=0,1,\cdots,199$,
$$(600-2k)(599-2k)(k+1)\lt 300^3$$
Using the lemma, 
$$\begin{align}600!&=\prod_{k=0}^{199}(600-2k)(599-2k)(k+1)\lt (300^3)^{200}=3^{600}\end{align}$$
Proof for lemma : 
$$(600-2x)(599-2x)(x+1)\lt 300^3$$
is equivalent to 
$$4 x^3-2394 x^2+357002 x-26640600\lt 0$$
Let $f(x)$ be the LHS. Then,
$$f'(x)=12 x^2-4788x+357002$$
Let $x=\alpha$ be the smaller solution of $f'(x)=0$. 
Then, noting that $\alpha\gt 0$, we have
$$f(\alpha)=\frac{2i-399}{6}f'(\alpha)-\frac{241202}{3}\alpha-2899967\lt 0$$
With $f(199)\lt 0$, we have $f(x)\lt 0$ for $0\le x\le 199$. $\blacksquare$
A: $$\ln{n!} = \sum_{k=1}^n\ln{k}$$
solve for $n!$ :
$$n! = e^{\sum_{k=1}^n\ln{k}}$$
approximate $ \sum_{k=1}^{600}\ln{k} \approx 3242.2753...$
$$600!\approx e^{3242.2753...}$$
On the other hand we can express $300^{600}$ as $e^{\ln{(300^{600})}}$. Then using logarithm rules:
$$300^{600}=e^{600\ln{300}}\approx e^{3422.2694...}$$
$$e^{3242.2753...}<e^{3422.2694...}$$
$$600! < 300^{600}$$
(P.S. This is one of my first posts, I'm sorry if it's not the best work.)
