Define the exponential Brownian martingale as $N_t = \exp\left\{a W_t - \frac12 a^2 t \right\}$ which is a martingale with respect to the natural filtration of $W$ which stands for a standard Brownian motion.
I now want to prove that $N_t \rightarrow 0$ a.s. for $t \rightarrow \infty$. As $N_t$ is a martingale it is by definition in $L^1$ hence Doob's martingale convergence theorem tells us that there exists a $N_\infty \in L^1$ such that $N_t \rightarrow N_\infty$ a.s., but how could I prove that $N_\infty \equiv 0$? Any help is greatly appreciated!
As @Did already pointed out, it is much easier to show that
$$a W_t - \frac{1}{2} a^2 t \to - \infty \qquad \text{almost surely as $t \to \infty$.}$$
Recall that the process $$B_t := \begin{cases} t W_{\frac{1}{t}}, & t>0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion; in particular
$$\lim_{t \to 0} t W_{\frac{1}{t}} = \lim_{t \to 0} B_t = 0.$$
This implies
$$\lim_{t \to \infty} \frac{W_t}{t}= \lim_{s \to 0} s W_{\frac{1}{s}} = 0.$$
Hence,
$$a W_t - \frac{1}{2} a^2 t = t \underbrace{\left( a \frac{W_t}{t} -\frac{a^2}{2} \right)}_{\to - \frac{a^2}{2}} \to - \infty$$
almost surely as $t \to \infty$.