Let's prove that $(100!)^{99}>(99!)^{100}$
$$\frac{(100!)^{99}}{(99!)^{99}}=100^{99}>99!=\frac{(99!)^{100}}{(99!)^{99}}\implies (100!)^{99}>(99!)^{100}$$
Let's prove that $(99!)^{100}>(100!)^{98}$
$$\sqrt{\frac{(99!)^{100}}{(99!)^{98}}}=99!>100^{49}=\sqrt{100^{98}}=\sqrt{\frac{(100!)^{98}}{(99!)^{98}}}\implies (99!)^{100}>(100!)^{98}$$