Perfect number from Euclid's formula

Statement

Euclid’s formula says that if $2^p-1$ is prime, then $N=2^{p-1}\,(2^p-1)$ is a perfect number. Compute the perfect number for $p=7$.

Solution

With $p=7$: $2^7-1=127$, which is prime. Hence $N=2^{6}\cdot 127=64\cdot 127=\boxed{8128}.$