If there is an n^(100^100) algorithm that solves an NP-complete problem, then P=NP, but public-key cryptography is still safe because for any practical n it's s

If there is an n^(100^100) algorithm that solves an NP-complete problem, then P=NP, but public-key cryptography is still safe because for any practical n it's still too hard to break. There are also public-key systems that are based on NP-complete problems that are easily broken, because n is chosen too small.