This may not be exactly what you're looking for, but here's a proof which at least uses the specific form of the process (being the exponential local martingale of an integral of a square-integrable deterministic process with respect to a Brownian motion):
First off, in order to make things fit better with the standard framework, I'll assume that $\phi$ is a mapping from $[0,\infty)$ to $\mathbb{R}$ with the property that $\int_0^t \phi(s)^2ds$ is finite for all $t\ge0$. Define $N_t = \int_0^t \phi(s)dW_s$, we then have $M_t = \mathcal{E}(N)_t$, the exponential local martingale, and the objective is to prove that $\mathcal{E}(N)$ is a martingale (instead of just a local martingale). By some classical results, $\mathcal{E}(N)$ is a nonnegative supermartingale, and it is a martingale if and only if $E\mathcal{E}(N)_t = 1$ for all $t\ge0$. I don't know where you actually can find a proof of these claims, but they follow from applications of the optional sampling theorem and Fatou's lemma. The conclusion is that we need to show $E\mathcal{E}(N)_t=1$ for all $t\ge0$.
To do so, first note that $[N]_t = \int_0^t \phi(s)^2 ds$. We apply Itô's formula:
$\mathcal{E}(N)^2_t = 1 + 2\int_0^t \mathcal{E}(N)_sd\mathcal{E}(N)_s+[\mathcal{E}(N)]_t\\
=1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_sd[N]_s\\
=1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_s\phi(s)^2ds.$
As $\mathcal{E}(N)$ and $\mathcal{E}(N)\cdot \mathcal{E}(N)$ are continuous local martingales, they are locally bounded. Let $(T_n)$ be a localising sequence of stopping times such that both $\mathcal{E}(N)^{T_n}$ and $(\mathcal{E}(N)\cdot \mathcal{E}(N))^{T_n}$ are bounded martingales. By the martingale property of $(\mathcal{E}(N)\cdot\mathcal{E}(N))^{T_n}$, we then obtain
$E\mathcal{E}(N)_{t\land T_n}^2 = 1 + E\int_0^{t\land T_n}\mathcal{E}(N)_sd\mathcal{E}(N)_s
+ E\int_0^{t\land T_n}\mathcal{E}(N)^2_s\phi(s)^2 ds\\
=1 + E \int_0^t \mathcal{E}(N)^2_s\phi(s)^21_{[0,T_n]}(s)ds
\le 1 + \int_0^t E\mathcal{E}(N)^2_{s\land T_n}\phi(s)^2ds$
where we also applied Tonelli's theorem and nonnegativity of $\mathcal{E}(N)$. Now consider a fixed $n\ge1$ and define $g_n(t) = E\mathcal{E}(N)^2_{t\land T_n}$. The above then states that
$g_n(t) \le 1+ \int_0^t g_n(s)\phi(s)^2ds$.
By a classical analysis lemma, Gronwall's lemma (See the Wikipedia article on Gronwall's inequality), we then obtain $g_n(t) \le \exp(\int_0^t\phi(s)^2ds)$. In other words, we have now shown
$E\mathcal{E}(N)^2_{t\land T_n}\le \exp(\int_0^t \phi(s)^2ds)$
for all $n\ge1$ and $t\ge0$. Now fix $t\ge0$. By the above, the family $(\mathcal{E}(N)_{t\land T_n})_{n\ge1}$ is then bounded in $\mathcal{L}^2$, therefore uniformly integrable. Furthermore, $\mathcal{E}(N)_{t\land T_n}$ converges almost surely to $\mathcal{E}(N)_t$. Combining this with uniform integrability, $\mathcal{E}(N)_{t\land T_n}$ converges in $\mathcal{L}^1$ to $\mathcal{E}(N)_t$, and so the means also converge. And as $\mathcal{E}(N)^{T_n}$ is a martingale, $\mathcal{E}(N)_{t\land T_n}=1$. We conclude that $E\mathcal{E}(N)_t = 1$, and so $\mathcal{E}(N)$ is a martingale.
In order to make sense of a stochastic integral of the form $\int_0^t f(B_s) \, dB_s$ you need to verify two properties:
- $f$ is suitably measurable, e.g. progressively measurable
- $f$ is suitably integrable, e.g. $\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$ or $\mathbb{E}(\int_0^{t \wedge \tau_n} f(s)^2 \, ds)<\infty$ for a sequence of stopping times $\tau_n \uparrow \infty$
Under the above conditions, the stochastic integral $M_t := \int_0^t f(B_s) \, dB_s$ is a local martingale but, in general, it might fail to be a martingale. In order to ensure that $(M_t)_{t \geq 0}$ is a martingale (not only a local one), $f$ has to satisfy the stronger integrability condition of the above-mentioned conditions, that is, $\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$ for all $t\geq 0$.
This is exactly the condition which the authors of the solution verified for the particular case $f(s):=B_s^2$.
Best Answer
Itô's formula shows that
$$M_t = 1+ \int_0^t f(s) M_s \, dW_s \tag{1}$$
and this implies, in particular, that $(M_t)_{t \geq 0}$ is a local martingale. (Note that $(M_t)_{t \geq 0}$ has continuous sample paths, and therefore the stochastic integral on the right-hand side is well-defined.) On the other hand, it follows from the very definition that $M_t \geq 0$ for each $t \geq 0$. Since any non-negative local martingale is a supermartingale (see e.g. this question for details), we conclude that $(M_t)_{t \geq 0}$ is a supermartingale. Thus,
$$\mathbb{E}(M_t) \leq \mathbb{E}(M_0) \leq 1$$
which implies
$$\mathbb{E} \exp \left( \int_0^t f(s) \, dW_s \right) \leq \exp \left( \frac{1}{2} \int_0^t f(s)^2 \, ds \right)$$
for each $t \geq 0$. Replacing $f$ by $2f$ we find in particular that
$$\mathbb{E} \left| \exp \left( \int_0^t f(s) \, dW_s \right) \right|^2 = \mathbb{E}\exp \left( \int_0^t 2f(s) \, dW_s \right) \leq \exp \left( 2 \int_0^t f(s)^2 \, ds \right) < \infty,$$
and so
$$\mathbb{E}(M_t^2) \leq \exp \left( \int_0^t f(s)^2 \, ds \right).$$
Using this estimate and the fact that $f$ is deterministic, we can easily check that the stochastic integral $\int_0^t f(s) M(s) \, dW_s$ is a true martingale, and now $(1)$ shows that $(M_t)_{t \geq 0}$ is a martingale.