If $a=1$ then trivial else denote $f(b)=ab-b\ln{b}$ for $b\in[1,\infty)$, then $f'(b)=a-\ln{b}-1=0\Longleftrightarrow b=e^{a-1}\in (1,\infty)$ and $f(e^{a-1})$ is maximum of $f$. But $f(e^{a-1})=ae^{a-1}-(a-1)e^{a-1}=e^{a-1}$. Done!