Expected stopping time Let $X_1,X_2,\ldots,$ be iid uniform in $[0,1]$ and define $S_n:=\sum_{i=1}^n X_i$. Let $\tau$ be the smallest $n$ for which $S_n>1$. It is known that $E[\tau]=e\approx 2.7183$, and this can be easily shown using the
Irwin-Hall https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution 
distribution. Indeed,
$$ P(\tau\ge n) = P(S_{n-1}< 1)=1/(n-1)!,$$
whence
$$ E[\tau]=\sum_{n=1}^\infty P(\tau\ge n) = \sum_{n=0}^\infty 1/n! = e.$$
Here, all of the heavy-hitting is done by the
Irwin-Hall distribution, and I'm wondering: is there a clever "soft" martingale argument that gives the answer with less heavy-hitting?

One way, although it does not use martingales, is to consider a discrete time Markov process whose state space is $[0,1] \cup \{ \square \}$ (the choice of symbol for the extra state is not important). This process says that $X_{n+1}$ conditioned on $X_n$ is equal to $X_n+u$ for $u \sim U([0,1-X_n])$ with probability $1-X_n$ and is equal to $\square$ with probability $X_n$. And $X_{n+1}$ conditioned on $X_n=\square$ is $\square$. Define 
$$(Pf)(x)=E[f(X_{n+1}) \mid X_n=x]$$
Explicitly:
$$(Pf)(x)=\int_x^1 f(y) dy + x f(\square) \quad x \in [0,1] \\
(Pf)(\square)=f(\square).$$ 
Then define $L=P-I$. Renewal theory tells you that if $\tau$ is the time to hit $\square$ then $E[\tau \mid X_0=x]$ satisfies the equation
$$(Lu)(x)=-1 \quad x \in [0,1] \\
u(\square)=0$$
which expands out to
$$\int_x^1 u(y) dy - u(x) = -1 \quad x \in [0,1].$$
You want $u(0)$. Differentiating, you get $-u(x)-u'(x)=0$ so that $u=Ce^{-x}$. Plugging back into the integral equation you get $-\frac{C}{e}=-1$ so $C=u(0)=e$. Alternately you could note that $u(1)$ must be $1$ (why?) and get the same result.
When I say "renewal theory" this is not really something complicated; it is just saying that $E[\tau \mid X_0=x]$ is $E[\tau \mid X_1=y]$ integrated in $y$ against the conditional distribution of $X_1$ given $X_0=x$. Then the thing being integrated is the same as $1+E[\tau \mid X_0=y]$, where this $1$ takes into account the fact that a step takes $1$ time.
