IBM Ponder This #197 · September 2014

Let b = (2^3^4^5) / (e^n), otherwise stated as "two to the power of (three to the power of (4 to the power of 5)) over e to the nth power", where n is an integer such that

1 < b < e.

Find b, with an accuracy of 10 decimal digits.

Update (9/11): There is a rather elegant way to solve this problem. We've added a star to solvers who found it.

Solution

Best opened after a real attempt

To be added.