Stochastic exponential of a true martingale is also a true martingale?

Question

Let $M=(M_t)_{t\geq 0}$ be a true martingale. Is the stochastic exponential $(\mathcal{E}(M)_t)_{t\geq 0}$ i.e. \begin{equation} \frac{d\mathcal{E}(M)_t}{\mathcal{E}(M)_t}=dM_t, \quad \quad \mathcal{E}(M)_0=1 \end{equation} also a true martingale?

I know that $\mathcal{E}(M)_t$ is a supermartingale as it is bounded below by $0$.