Proof:
P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n}
(Bayes theorem)
=P{T>n+m}/P{T>n}
=((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m
(1-p)^m = {T>m}
So T>m has the same probability as T>m+n given that
T>n, which means it doesn't care (or don't remember) that n
phases had passed.