When n=1, the product is just one term 1-½=½. This is the largest value the product can be, because the other operands are less than 1, so their product must be smaller than ½.
When n=1, ¼+1/(2^(n+1))=¼+¼=½. This ties in with the result in the previous paragraph when n=1, and proves the base case for n=1.
As n approaches infinity, this right-hand expression (hypothesis) approaches ¼ because 1/(2^(n+1) approaches 0. So ¼<¼+1/(2^(n+1))≤½, according to the hypothesis. Let p[n]=¼+1/(2^(n+1)).
p[n+1]=p[n](1-1/(2^(n+1)) from the formula.
p[n+1]/p[n]=1-1/(2^(n+1))
1/(2^(n+1))=(p[n]-p[n+1])/p[n]
Assume the hypothesis p[n]≥¼+1/(p^(n+1)), 1/(p^(n+1))≤p[n]-¼
Therefore (p[n]-p[n+1])/p[n]≤p[n]-¼
1-p[n+1]/p[n]≤p[n]-¼
p[n+1]/p[n]≥5/4-p[n], according to the hypothesis.
p[n+1]/p[n]=1-1/(2^(n+1)) according to the formula.
So we have to prove 1-1/(2^(n+1))≥5/4-p[n]. But ¼<p[n]≤½, according to the hypothesis, so 5/4-p[n] has a maximum value of <1 and a minimum value of 3/4. 1-1/(2^(n+1)), according to the formula, has a maximum value of <1 and a minimum value of 3/4. The maximum and minimum values for the hypothesis and the formula coincide. Therefore by induction, QED.