10^n≡（-1)^n（mod11）
by MI
n=1
LS=10^1= 10= (-1)^1(mod11)
Assume p(k) is true
ie
10^k≡（-1)^k(mod11)
for n=k+1
LS
10^(k+1)
=10^k (10)
=[ （-1)^k(mod11) ].10
= [ （-1)^k+ 11m ].10 ( where m is +ve integer )
= (-1)^k.10 + 110m
= (-1)^k.10 (mod 11)
= (-1)^k ( 11-1) (mod 11)
= (-1)^(k+1) ( mod 11 )
By principle of MI it is true for all n