Find the Largest Number from 50^99 and 99!

Which is the largest number from 50^99 and 99!
Where 50^99 = 50 × 50 × 50 ×……..× 50 {99 terms}
and 99! = 99 × 98 × 97 ×……..× 1
Solution: Finding the largest number
Let start with 50^99/99!
\begin{aligned} \frac{50^{99}}{99!}&= \frac{50\times50\times50\times......\times50}{99\times98\times97\times......\times1}\\ \end{aligned}
Rearrange the right-handed side
\begin{aligned} \frac{50^{99}}{99!}&= \frac{50\times50}{99\times1}\times\frac{50\times50}{98\times2}\times\frac{50\times50}{97\times2}\times........\times\frac{50\times50}{51\times49}\times\frac{50}{50}\\ \\ &= \frac{50^2}{(50+49)(50-49)}\times\frac{50^2}{(50+48)(50-48)}\times........\times\frac{50^2}{(50+1)(50-1)}\times\frac{1}{1}\\ \end{aligned}
We know (a + b)(a – b) = a² – b²
so
\begin{aligned} \frac{50^{99}}{99!}&= \frac{50^2}{50^2-49^2}\times\frac{50^2}{50^2-48^2}\times........\times\frac{50^2}{50^2-1^2}\\ \end{aligned}
Here
\begin{aligned} \frac{50^2}{50^2-49^2}&>1 \\ \\ \frac{50^2}{50^2-48^2}&>1\\ &.\\ &.\\ &.\\ &.\\ &.\\ \frac{50^2}{50^2-1^2}&>1\\ \end{aligned}
So we can write
\begin{aligned} \frac{50^{99}}{99!}&>1\times1\times........\times1\\ \\ \frac{50^{99}}{99!}&>1\\ \\ 50^{99}&>99! \end{aligned}