Iterative squaring is not paralellizable - to compute x^4 as (x^2)^2 you have to compute t = x^2 first before starting the computation of t^2. Calculating x^(2^

Iterative squaring is not paralellizable - to compute x^4 as (x^2)^2 you have to compute t = x^2 first before starting the computation of t^2. Calculating x^(2^t) mod n takes t squarings.