It Telescopes? : Extreme Edition

Let's begin by noting the first few $f_{n}(x)$ .

$f_{0}(x) = e^{x}, \ f_{1}(x) = e^{e^{x}}, \ f_{2}(x) = e^{e^{e^{x}}}, \ f_{3}(x) = e^{e^{e^{e^{x}}}}, \ ...$

We can see that $f_{n}(x)$ is a power tower of $n+1$ $e$ 's with an $x$ topping them off. Let $P_{n}(x)$ be the product inside the integral. Then

$P_{n}(x) = \prod_{k=0}^{n} f_{k}(x) = e^{x}e^{e^{x}}e^{e^{e^{x}}} ... \underbrace{e^{e^{e^{...^{x}}}}}_{n+1 \text{ e's}}$

Notice that $\frac{d}{dx} \underbrace{e^{e^{e^{...^{x}}}}}_{n+1 \text{ e's}} = P_{n}(x)$ Then we can rewrite the integral as follows:

$\int_{0}^{1} P_{n}(x) dx = \int_{0}^{1} \frac{d}{dx} \underbrace{e^{e^{e^{...^{x}}}}}_{n+1 \text{ e's}} dx = \int_{0}^{1} d \left( \underbrace{e^{e^{e^{...^{x}}}}}_{n+1 \text{ e's}} \right) = \underbrace{e^{e^{e^{...^{1}}}}}_{n+1 \text{ e's}} - \underbrace{e^{e^{e^{...^{0}}}}}_{n+1 \text{ e's}} = \underbrace{e^{e^{e^{...^{1}}}}}_{n+1 \text{ e's}} - \underbrace{e^{e^{e^{...^{1}}}}}_{n \text{ e's}}$

Now, looking at $S$ we get

$S = \sum_{n=0}^{2018} \int_{0}^{1} P_{n}(x) dx = \sum_{n=0}^{2018} \underbrace{e^{e^{e^{...^{1}}}}}_{n+1 \text{ e's}} - \underbrace{e^{e^{e^{...^{1}}}}}_{n \text{ e's}}$

Which is a nice telescoping sum! The sum can then be easily be evaluated as $S = \underbrace{e^{e^{e^{...^{1}}}}}_{2019 \text{ e's}} - 1 = \underbrace{e^{e^{e^{...}}}}_{2019 \text{ e's}} - 1$ . The 1 has been dropped since the $e^{1}$ at the end of the power tower is simply equal to $e$ . Now note that

$\underbrace{e^{e^{e^{...}}}}_{2019 \text{ e's}} > \underbrace{e^{e^{e^{...}}}}_{2019 \text{ e's}}-1 >\underbrace{e^{e^{e^{...}}}}_{2018 \text{ e's}}$

And also note that since the natural log is strictly increasing over its domain, applying $\ln$ to all three preserves the inequality. Lastly, note that applying $\ln$ to any power tower of $e^{e^{e^{e^{...^{e}}}}}$ consisting of finitely many $e$ 's removes one from the power tower. Thus, applying $\ln$ 2018 times to both sides gives

$e > \underbrace{\ln(\ln(\ln(\ln(...\ln(S)...))))}_{2018 \text{ times}} > 1 \\ 1 > \underbrace{\ln(\ln(\ln(\ln(...\ln(S)...))))}_{2019 \text{ times}} > 0 \\ 0 > \underbrace{\ln(\ln(\ln(\ln(...\ln(S)...))))}_{2020 \text{ times}} > -\infty$

And so our answer is $\boxed{2020}$