= Solution
The <strong law of large numbers> gives $S_n/n\to-1$ almost surely, so $S_n\to-\infty$. Therefore the infinite-horizon maximum $Z$ is finite and attained almost surely; also $Z\geq0$ because $S_0=0$.
Apply the <Doob maximal inequality> to $M_k=e^{2S_k}$ through time $n$. Since $\mathbb E M_n=1$,
$$
\mathbb P\left(\max_{0\leq k\leq n}S_k>t\right)\leq e^{-2t}\qquad(t\geq0).
$$
These finite-horizon events increase to $\{Z>t\}$. Continuity of <probability> under increasing unions gives
$$
\boxed{\mathbb P(Z>t)\leq e^{-2t}=e^{-\lambda t}.}
$$
For $t<0$ the same bound is trivial, because its right side exceeds $1$. Only bounded-horizon maximal inequalities were used before passing to the limit; no unjustified stopping at an infinite-horizon crossing time is needed.
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