$$\frac{(100!-99!)^{100}-(99!-98!)^{100}}{(98!-97!)^{100}}$$
$$\frac{(99! 99)^{100}-(98! 98)^{100}}{(97! 97)^{100}}$$
i was able to solve only till. How to proceed after this ?
How to solve these type of questions quickly ? is there any easy method ? Please help me.
Thanks in advance.
$$\frac{(100!-99!)^{100}-(99!-98!)^{100}}{(98!-97!)^{100}}=98^{100}\frac{(100\cdot99-99)^{100}-(99-1)^{100}}{(98-1)^{100}}=\left(\frac{98}{97}\right)^{100}(99^{200}-98^{100}).$$