I haven't worked this out, but this might lead to a proof -
Consider a cutoff value x0.
Calculate the probability that all are less than x0. You can calculate this from integrating the pdf to find the probability that one is less than x0 and taking the n'th power (everything is independent).
Now calculate the probability that all are less than x0. Subtract this probability from 1. This is the probability that some exceeds x0.
When you compare the 2 expressions, you should see that it's likely your relation holds.
What's left is figuring out how the value of n depends on x0 and your 0.99 value.