Second attempt. 2^n has approximately n log 2 digits. Assuming that they are distributed randomly, the probability that none of these digits are 0 is (9/10)^(n

Second attempt. 2^n has approximately n log 2 digits. Assuming that they are distributed randomly, the probability that none of these digits are 0 is (9/10)^(n log 2). For n = 50,000, this is already unbelievably tiny. For n = 50,001, it is even tinier. And so on. In fact, summing all of these tiny probabilities from n = 50,000 to n = infinity is about the same as integrating (9/10)^(x log 2) from x = 50,000 to x = infinity, which is, again, only slightly above zero. So, statistically at least, there is an excellent chance that every power of 2 above 50,000 has at least one zero.