Staying Positive

Using the properties of factorials, we know that $\dfrac{(n+1)!}{n!} = n+1$ for positive integer $n$ . So,

$\begin{aligned} \dfrac{100!}{99!} &=& 100 \\ \dfrac{100!}{99!\times 5!} &= &\dfrac{100}{5!} = \dfrac{100}{120} < 1 \\ 100! &< & 99!\times 5! \\ 100! -99! \times 5! &< &0 . \\ \end{aligned}$

$100! = (99!)(100)$

$(99!)(5!) = (99!)(120)$

$100 < 120$

Thus,

$(99!)(100) < (99!)(120)$

$(99!)(100) - (99!)(120) < 0$

$100!-99!\cdot 5! = 100\cdot 99!-99!\cdot 5! = 99!(100-5!) = 99!(100-120) = -1\cdot (20\cdot 99!)$ which is negative so less than 0.

Or just expand $5!$ and separate the $100$ from the $100!$ to get $100\cdot 99!-120\cdot 99!$ which is clearly negative.