The number you computed is nearly 1/e. Basically,
(1-1/n)^n = exp(n log (1-1/n))
= exp(n (-1/n + 1/(2n^2) + O(1/n^3) ))
= exp(-1 + 1/(2n) + O(1/n^2) )
= 1/e - 1/(2en) + O(1/n^2)
and so your answer should agree with 1/e to, say, the first forty digits or so.
(Incidentally, Maple choked when I plugged in (1-2^-128)^(2^128), because it tried to evaluate it as a rational number, the numerator and denominator of which would have well over 2^128 digits.)
(Incidentally, Maple choked when I plugged in (1-2^-128)^(2^128), because it tried to evaluate it as a rational number, the numerator and denominator of which would have well over 2^128 digits.)