Is this martingale identically zero? Let $X_t$ be such that $X_t$ is bounded continuous martingale adapted to the filtration $\mathcal{F}_t$ such that 
$$\Bbb{E} \bigg[\int_0^t  e^{X_s} \, d\langle X\rangle_s\bigg] = 0$$
and $X_0=0$. Does it follow that $X_t$ is identically zero?
remark: This question came from the following 
 $$\forall \lambda >0 ; e^{\lambda X_t} \text{ is a martingale} \Rightarrow X_t \equiv 0$$
I see two ways to solve this:
0) $\Bbb{E}[e^{\lambda (X_t - X_s)}\vert \mathcal {F}_s] = 1$ therefore
$$\Bbb{E}[e^{\lambda (X_t - X_s)} - 1\vert \mathcal {F}_s] = 0\\
\Bbb{E}\bigg[\frac{e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0\\
\lim_{\lambda\to 0}\Bbb{E}\bigg[\frac{e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0\\
\Bbb{E}\bigg[\frac{\lim_{\lambda\to 0}e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0 \\
\Bbb{E}[X_t - X_s\vert \mathcal {F}_s] = 0$$
therefore $X_t$ is a martingale. 
A) Since the exponential transform (extend it to complex you get characteristic functions, or just consider the laplace transform) allows one to recover the distribution, we get that $$\Bbb{E}[e^{\lambda X_t}] = 1 = \Bbb{E}[e^{\lambda 0}] $$
B) by Itô 
$$e^{X_t} = 1 + \int_0^t e^{X_s} \, dX_s + \int_0^t e^{X_s}\, d\langle X\rangle_s $$ take expectations
$$\Bbb{E}[e^{X_t}] = 1 +  \int_0^t \Bbb{E}[e^{X_s}]\, d\langle X\rangle_s $$
So basically I am asking how to proceed along the lines of $B)$
 A: Recall that for any function $f \geq 0$ and any measure $\mu$ we have
$$\int f \, d\mu = 0 \implies f=0 \quad \mu\text{-a.s.} $$
Therefore, as $(\langle X \rangle_t)_{t \geq 0}$ is an increasing process,
$$\mathbb{E} \left( \int_0^t e^{X_s} \, d\langle X \rangle_s \right)=0$$
implies
$$\int_0^t e^{X_s} \, d \langle X \rangle_s =0$$
$\mathbb{P}$-almost surely. Since $e^x$ is strictly positive and $(\langle X \rangle_t)_{t \geq 0}$ an increasing continuous process, this already implies
$$\langle X \rangle_t = \text{const} = \langle X \rangle_0 = 0$$
almost surely. Indeed: Suppose that it is not constant. Then we can pick $t_0 >0$ such that $\langle X \rangle_{t_0}(\omega) > \langle X \rangle_{0}(\omega)$. By the very definition of the Riemann-Stieltjes integral and the strict positivity of $e^x$, this implies
$$\int_{0}^{t_0} e^{X_s(\omega)} \, d\langle X \rangle_s(\omega)>0$$
in contradiction to our assumption.
Consequently,
$$\mathbb{E}(X_t^2) = \mathbb{E}(\langle X \rangle_t) = 0$$
which implies $X_t = 0$ almost surely.
