If $A^{\wedge}$ is the completion of a local ring, do we always have $\mathfrak{m}^{\wedge}=\mathfrak{m}A^{\wedge}$?

Question

Let $A$ be a local ring with maximal ideal $\mathfrak{m}$, and let $A^{\wedge}$ be the $\mathfrak{m}$-adic completion of $A$. Then $A^{\wedge}$ is a local ring with maximal ideal $\mathfrak{m}^{\wedge}=\ker(A^{\wedge}\to A/\mathfrak{m})$. In the case of the $p$-adic numbers, we have $$ \mathfrak{m}^{\wedge}=\mathfrak{m}A^{\wedge}. $$ Does this equality always hold?