What is the expectation of $\prod\limits_{k=1}^\infty \left(1+\prod\limits_{i=1}^k u_i\right)$, where the $u$'s are i.i.d. $\text{Uniform}(0,1)$-variables ?
My first thought was to try replacing $u_i$ with its mean value $\frac12$, giving $\prod\limits_{k=1}^\infty \left(1+\left(\frac12\right)^k\right)=$ 2.38423102903..., but this approach did not seem justifiable.
Taking the log of the products didn't seem to help.
An Excel simulation with half a million trials gave an average (arithmetic mean) of $2.7201$, suggesting that the answer is just $e$.
Context: This question was inspired by a related question asking for the probability that $\sum\limits_{k=1}^\infty \left(\prod\limits_{i=1}^k u_i\right)>1$.
Write
$$ f(x)=\mathsf E_{u_1,u_2,u_3,\ldots}[(1+xu_1)(1+xu_1u_2)(1+xu_1u_2u_3)\cdots]\;. $$
You want $f(1)$. We have
\begin{eqnarray} f(x) &=& \int_0^1\mathsf E_{u_2,u_3,\ldots}[(1+xu_1)(1+xu_1u_2)(1+xu_1u_2u_3)\cdots]\,\mathrm du_1 \\ &=& \int_0^1(1+xu_1)f(xu_1)\,\mathrm du_1 \\ &=& \frac1x\int_0^x(1+t)f(t)\,\mathrm dt \end{eqnarray}
(with the substitution $t=xu_1$). Multiplying by $x$ and then differentiating with respect to $x$ yields
$$ f(x)+xf'(x)=(1+x)f(x)\;, $$
and then cancelling $f(x)$ and dividing by $x$ yields
$$ f'(x)=f(x)\;, $$
with the solution $f(x)=c\,\mathrm e^x$. Since $f(0)=1$, we have $c=1$ and thus $f(1)=\mathrm e$.