Volterra product integral identity (proof) I am reading myself about Product integrals and I find three types of definition. See, for instance, https://en.m.wikipedia.org/wiki/Product_integral 
I am interested on the type I product integral. For the scalar case, $f:[a,b]\to\mathbb{R}$, it is defined
$$\prod_a^b (1+f(x)dx) :=\lim_{n\to \infty} \prod_{i=1}^n (1 + f(x_i)\Delta x_i)$$
where $P=\lbrace x_0,...,x_n\rbrace$ is a partition of $[a,b]$ and $\Delta x_i = x_{i} - x_{i-1}$.
I wonder why
$$ \lim_{n\to \infty} \prod_{i=1}^n (1 + f(x_i)\Delta x_i) =  \exp \left( \int_a^b f(x)\ dx \right)? $$
My idea was, assuming that the limit $\lim_{n\to \infty} \prod_{i=1}^n (1 + f(x_i)\Delta x_i) = L$ exists, taking logarithm and get
$$ \log \left( \lim_{n\to \infty} \prod_{i=1}^n (1 + f(x_i)\Delta x_i) \right) = \lim_{n\to \infty} \log \left( \prod_{i=1}^n (1 + f(x_i)\Delta x_i) \right) = \lim_{n\to\infty} \sum_{k=1}^n \log (1+f(x_i)\Delta x_i).$$
But now what? I am stucked here. Any advice or help is welcome. Thanks in advance.
 A: Continuing from where you were stuck.
The reason is that for very small $\Delta x_i$, $\log (1+f(x_i)\Delta x_i) \approx f(x_i)\Delta x_i$.
This is because the taylor expansion around 1 of $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdot \cdot \cdot$
So,
$\sum_{k=1}^n \log (1+f(x_i)\Delta x_i) \to \sum_{k=1}^n f(x_i)\Delta x_i $ as $n \to \infty$
and that last expression is just the Riemann sum definition of the integral.
Sometimes you see Riemann sum defined with a shrinking $\Delta x_i$, other times you see it with a growing $n$ in the denominator. Roughly, $\Delta x_i = \frac{1}{n}$. So, equivalent to the fact about $\log$, for $n$ much larger than $f(x_i)$,
$$\left(1 + \frac{f(x_i)}{n}\right) = \left[\left(1 + \frac{f(x_i)}{n}\right)^n\right]^{\frac{1}{n}} \approx \left[\exp(f(x_i))\right]^{\frac{1}{n}} = \exp\left(\frac{f(x_i)}{n}\right)$$
since as $n \to \infty$, the part inside the square brackets approaches $\exp(f(x_i))$, by definition of $\exp$.
and this is also equivalent to the fact that the first term in the taylor expansion of $\exp(x)$ around 0 is $x$.  It makes sense that $\exp$ and $\ln$ are approximately linear (with slope of 1) at 0 and 1 respectively, since they're inverses of each other.
That's kind of how $e$ or $\exp$ tends to arise.  We approach the integral when we have sums of very man numbers near zero.  We approach the product integral, or $\exp(something)$ when we have products of very many numbers near 1.
$\sum \to \int$
$\prod \to \exp\left(\int\right)$
We're often introduced to $e$ with continuously compounded interest.  But what if the interest rate itself varied as a function of time, $f(t)$?  That's exactly where we would see the product integral.
Rather than thinking of the product integral as somehow being built with $e$, maybe it should be the other way around -- i.e., think of $e$ as the product integral of the function $f(x) = 1$ on the interval $[0, 1]$.
