why does (3/4)^-2 gives 16/9? Isn't it suppose to be 9/16?

$${\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{\left(-{\mathtt{2}}\right)} = {\left({\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{{\mathtt{2}}}\right)}^{\left(-{\mathtt{1}}\right)}$$

$${\left({\frac{{\mathtt{9}}}{{\mathtt{16}}}}\right)}^{-{\mathtt{1}}} = {\frac{{\mathtt{1}}}{\left({\frac{{\mathtt{9}}}{{\mathtt{16}}}}\right)}}$$

$${\frac{{\mathtt{1}}}{\left({\frac{{\mathtt{9}}}{{\mathtt{16}}}}\right)}} = {\frac{{\mathtt{16}}}{{\mathtt{9}}}}$$

Remember......when we have this situation (a/b)^(-m).....where a/b ≠ 0...we can  "flip"  the fraction and make the expononent positive.....so......

(3/4)^(-2)   =     (4/3)^2  =   (16/9)