Let $W = \{W_t : t\ge0\}$ be a standard Brownian motion on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $f$ be a deterministic function such that $$ \int_0^tf^2(s)\,ds<\infty $$ for all $t\ge 0$. Show that the stochastic exponential $$ M_t = \exp\left(\int_0^tf(s)\,dW_s - \frac{1}{2}\int_0^tf^2(s)\,ds\right) $$ is a martingale.
I'm aware that this can be verified immediately using Novikov's criterion, for example, but I'm looking for a more direct proof than this. It's easy to demonstrate using Ito's lemma that $$ M_t = 1 + \int_0^tf(s)M_s\,dW_s, $$ and so the result can be boiled down to showing that $$ \mathbb{E}\left[\int_0^tf^2(s)M_s^2\,ds\right]<\infty. $$ This seems promising, but I'm unsure how to proceed. I should mention that the subsequent part of the exercise I'm trying to solve asks to use the martingale property of $M_t$ to show that $$ \int_0^tf(s)dW_s\sim N\left(0,\int_0^tf^2(s)\,ds\right). $$ This is straightforward to do, but it suggests that the martingale property can be demonstrated without appealing to these facts. Any suggestions or references would be greatly appreciated.
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
$$\mathbb{E} \exp \left( \int_0^t f(s) \, dW_s \right) \leq \mathbb{E} \exp \left( \frac{1}{2} \int_0^t f(s)^2 \, ds \right)$$
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.