My plan is to confirm is that $M_t$ is a martingale. If true, then $E[M_{t+1}] = E[M_{t}]$, and since $E[M_0]=1$, then $E[M_t]=1$ for all $t$. We can then conclude that $$||M_t||_1=E[|M_t|]=1$$ $$\lim_{t \to \infty}|M_t - 0|=1$$ So $M_t$ doesn't converge to zero in $L^1$ norm.
From my book I have the following direction for the proof that $M_t$ is in fact a Martingale.
For $t > s$ we have $$E[M_t|\mathcal{F}_s] = E[\exp(\alpha (B_t - B_s + B_s) - \alpha^2 (t-s+s)/2)|\mathcal{F}_s]$$ $$= E[\exp(\alpha (B_s) - \alpha^2 (s)/2)\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]$$ $$= M_sE[\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]$$
The last step, which is not explained is that, somehow $E[\exp(\alpha (B_t - B_s) - \alpha^2 (t-s)/2)|\mathcal{F}_s]=1$, so we end up with $E[M_t|\mathcal{F}_s]=M_s$.
I think the idea is that $B_t - B_s = B_{t-s}$, so $\exp(\alpha B_{t-s})$ must equal $\exp(\alpha^2(t-s)/2)$, but I'm not quite sure how?
Note that $B_t-B_s\sim N(0,t-s)$. The moment generating function of a random variable $X$ with normal distribution $N(\mu,\sigma^2)$ is given by \begin{align} \mathbb{E}\left[e^{tX}\right]= e^{\mu t+\frac{1}{2}\sigma^2t^2} \end{align} Using these two observations (and the fact that Brownian motion has independent increments) we conclude that \begin{align} \mathbb{E}\left[e^{\alpha (B_t-B_s)}|\mathcal{F}_s\right]=e^{\frac{1}{2}(t-s)\alpha^2} \end{align}