Use only πs to get e

Question

I am taking this puzzle from inspiration from the Four Fours to get Pi puzzle. The puzzle rules are just slightly different. Here are the rules:

You are trying to get as close to e as possible with only πs used as real numbers in the approximations, and with the following allowed operations:

• Simple arithmetic: +, -, x, ÷
• Exponents and logs as binary operations: $\log_a b$ and $a^b$. You can only take logs of positive numbers.
• The root function as a binary operator $\sqrt[b]{a}$.
• Unary operations: $(\cdot)!$ for integer arguments only, unary negation -, the square root $\sqrt{\cdot}$ for positive numbers, and floor and ceil functions: $\lfloor\cdot\rfloor$ and $\lceil \cdot\rceil$.

Any other operations are not allowed, including double factorials and decimal points. In addition, you can do the following with no penalty:

• Parentheses (for grouping purposes only, no binomial coefficients etc.)

The score will be calculated the same way as in the four fours puzzle*.

Your score is equal to the number of digits of accuracy per operation used. That is, if you got the approximation A by using n operations, your score is

$$\frac{-\log_{10}|e - A|}{n}$$ (adjusted for e instead of π.)

*However, as a side note, you multiply your score by π/n where n is the number of πs used in your approximation.

The one with the highest score will get that sweet green checkmark on their submission.

Hint:

Try to use $$\lim_{n \to +\infty} \left( 1 + \frac{1}{n}\right)^n = e. $$

Answer 1

One can get

an arbitrarily high finite score

by

using the suggested formula $e\simeq\left(1+\frac1n\right)^n$ and taking $n=\pi^{\pi^{\pi^\pi}}$ with an arbitrarily large tower of $\pi$s. (Grouped so that you do the "upper" exponentiations first, of course.)

So the only question is

whether one can get an infinite score by hitting $e$ exactly. Obviously we could do it if allowed complex logarithms (with some care being needed about exactly how they're interpreted) since $e^{i\pi}=-1$, but I think it can't be done with the restrictions given in the question.