Textbook Problem

It might seem that $n-1$ multiplications are needed to compute $x^n$ , since

$x_{{}^n}=\underset{n-1\text{multiplications}}{\underset{︸}{x\cdot x\cdot \cdots x}}$

But observe that, for instance, since $6=4+2,x^6=x^4{x}^2={\left(x^2\right)}^2{x}^2$ . Thus $x^6$ can be computed using three multiplications: one to compute $x^2$, one to compute ${\left(x^2\right)}^2$ ,