$2^{60}$ when divided by $7$leaves the remainder

We have,

$\begin{array}{l} 2^{60}=(1+7{)}^{20}\\ \Rightarrow 2^{60}{=}^{20}{C}_0{+}^{20}{C}_1\times 7{+}^{20}{C}_2\times 7^2+\dots {+}^{20}{C}_{20}\times 7^{20}\\ \Rightarrow 2^{60}=7\left({ }^{20}{C}_1{+}^{20}{C}_2\times 7+\dots {+}^{20}{C}_{20}\times 7^{19}\right)+1\end{array}$

Hence, $2^{60}$ when divided by $7$leaves the remainder $1$.