# squaring binary doubles the nr of bits?

If a is a 10 bit number, then is $$|a^2|=20?$$ And then $$|a^3|=30?$$Or is that at least an upper bound? Does it work this way? I want to explain how with very high probability 4096 bit number is bigger than a 256 bit number to the power of 3. I think that 256 bit number to the power of 3 must be at most 768 bits, but could someone confirm it for me? Couldn't find any source stating that.

An $n$-bit number $a$ is a number with $2^{n-1}\le a<2^n$. Then for $a^k$ we have $2^{kn-k}\le a<2^{kn}$, i.e. $a^k$ has at least $kn-k+1$ and at most $kn$ binary digits. Especially the cube of a 256 bit number has between 766 and 768 bits.