Stupid question of the week. My book says "notice each number in

$\displaystyle b,b^2,{b^2}^2,{b^2}^3,{b^2}^4\ldots$

is the square of the preceding one"

In other words

$\displaystyle {b^2}^3={({b^2}^2)}^2$.

Fortunately,

$\displaystyle {({b^2}^2)}^2=({b^2}^2)({b^2}^2)={b^{2^2+2^2}={b^{ 2 \cdot2^2}={b^2}^3$. \checkmark (or \checkbox, or...whatEVER).

But also

$\displaystyle {({b^2}^2)}^2={(b^4)}^2=(b^4)(b^4)=b^8$.

I mean, two squaredisusually four, right?

Isn't $\displaystyle {b^2}^3={({b^2}^2)}^2$? (Actually, I know the answer: Yes.)

So what's wrong with my "But also..." line?