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$$\frac{(100!-99!)^{100}-(99!-98!)^{100}}{(98!-97!)^{100}}$$

$$\frac{(99! 99)^{100}-(98! 98)^{100}}{(97! 97)^{100}}$$

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.

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$$\frac{(100!-99!)^{100}-(99!-98!)^{100}}{(98!-97!)^{100}}=98^{100}\frac{(100\cdot99-99)^{100}-(99-1)^{100}}{(98-1)^{100}}=\left(\frac{98}{97}\right)^{100}(99^{200}-98^{100}).$$

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