Checking approximation of E

Question

MathWorld page gives a simple numeric formula for e that's allegedly correct for first 10^25 digits. It states that e is approximately

(1 + 9^-4^(7*6))^3^2^85

Any idea how to check whether this formula is correct even for the first 10 digits? Here's another way of writing the right hand side

Power[Plus[1, Power[9, Times[-1, Power[4, Times[7, 6]]]]], Power[3, Power[2, 85]]]

Answer 1

This problem does not need Mathematica at all. First, it is easy to show that 9^(4^(7*6)) is exactly equal to 3^2^85, since

 9^(4^(7*6)) = 3^(2*4^(7*6)) = 3^(2^(1+2*(7*6))) = 3^2^85

Then, we know that one of the ways to represent e is as a limit

e = lim (1+1/n)^n, n->infinity

The only question is what is the error given that n is very large but finite. We have

(1+1/n)^n = e^log((1+1/n)^n) = e^(n*log(1+1/n)) = e^(1-1/(2n)+O(1/n^2)) = e + O(1/n),

Given the n = 3^2^85, i we take the log(10,n) = 2^85 log(10,3) ~ 1.85 *10^25, we get an estimate similar to the quoted one