Simple Laws of exponents - that aren't so simple

Question

Evaluate $(b^2)^{\frac{1}{2}}$ for $b=-1$.

If you substitute $b=-1$ first and then simplify you get $((-1)^2)^{\frac{1}{2}}=(1)^{\frac{1}{2}}=1$ but if you simplify first and then substitute you get $b^1=(-1)^1=-1$.

What am I missing?