Prove analytically that for any nonnegative integer k there exists a power of 3, terminating with 0..001,
i.e. a sequence of k zeroes followed by digit 1,

e.g. k=0==> 3^0=1, 3^4 =81;

k=1==> 3^20=3486784401,

k=2==> 3^100= 515377520732011331036461129765621272702107522001.

Bonus question: Show that replacing the number 3 by any odd number, not terminating by 5, leaves
your proof valid.