www.mikescher.com/www/statics/euler/Euler_Problem-063_explanation.md

After all the previous programs, this one is surprisingly ... dense (the main code-block is 54x8).

The algorithm is quickly explained for each length n we calculate the numbers 1^n, 2^n ... until 9^n and see which have a length of n. (From 10^n upwards the condition is impossible, because 10^n has (n+1) digits).

The main problem is that the numbers exceed Int64. So we need to implement long multiplication ... again. (see problem 16, 20, 29, 56 and 57)