 Solution

The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.