phel

The last two digits are 99.

productfor(n, 0, 99, (99^n)) mod 10^10==7087255001 - these are the last 10 digits.

$\displaystyle \prod_{n=0}^{99} 99^n$

$99^{4950}$

As the guest before mentioned the last 2 digits are 01

-Vinculum

$99^0*99^1*99^2* .....*99^{99}\\ =99^{0+1+2+3+4+...99}\\ =99^{\text{an even number}}$

99^1=99

99^2=9801

99^3=ends in 99

99^4 ... ends in 01

etc

99 ^ (any even positive integer) will end in 01