= Solution
Here $X_1\leq1$, so for every $\tau>0$,
$$
0<e^{\tau X_1}\leq e^\tau,\qquad M(\tau)=\mathbb E e^{\tau X_1}\leq e^\tau<\infty.
$$
The <Jensen inequality> for the strictly convex <exponential function> gives
$$
M(\tau)\geq e^{\tau\mathbb E X_1}=1.
$$
In fact the inequality is strict: $\mathbb P(X_1=1)>0$ and $\mathbb E X_1=0$ exclude a constant <random variable>. In particular, $\boxed{1\leq M(\tau)<\infty}$ as required.
For $Z_n=e^{\tau S_n}/M(\tau)^n$, independence and identical distribution give
$$
\mathbb E e^{\tau S_n}=\prod_{i=1}^n\mathbb E e^{\tau X_i}=M(\tau)^n.
$$
Thus $Z_n$ is positive, integrable with mean one, and adapted to the <natural filtration>. Conditioning on the next increment gives
$$
\mathbb E[Z_{n+1}\mid\mathcal F_n]
=\frac{e^{\tau S_n}}{M(\tau)^{n+1}}\mathbb E[e^{\tau X_{n+1}}\mid\mathcal F_n]
=\frac{e^{\tau S_n}}{M(\tau)^n}=Z_n.
$$
Consequently $\boxed{(Z_n)\text{ is a martingale}.}$ This is the <exponential martingale of a random walk>.
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