Question

The last digit of ${17}^{256}$ is

Answer 1

We have,
${17}^{256}=({17}^2{)}^{128}=(289{)}^{128}=(290-1{)}^{128}={}^{128}{C}_0(290{)}^{128}-{}^{128}{C}_1(290{)}^{127}+\dots -{}^{128}{C}_{127}(290)+1=10k+1,k\in Z$

Hence, the last digit is $1$.