Ladder Of Magnitude

This is a problem where I like to find a pattern before formalizing.

Note that $10^{99} \div 10^{100} = 10^{-1}.$

Then, $10^{98} \div 10^{-1} = 10^{99}.$

Then, $10^{97} \div 10^{99} = 10^{-2}.$

Then, $10^{96} \div 10^{-2} = 10^{98},$ and so on.

Continuing this pattern, we will get $10^{-50}$ for the entire result.

To formalize the solution, we can set $a_1 = 10^{99} \div 10^{100} = 10^{-1}$ and $a_k = 10^{100-k} \div a_{k-1}$ for $k=2,3,\ldots,99.$ Using induction , we can prove the pattern we saw above.