Answer
$$K = \frac{{n\left( {n - 1} \right)}}{{{{\left( {1 + {n^2}} \right)}^{3/2}}}},{\text{ }}r = \frac{{{{\left( {1 + {n^2}} \right)}^{3/2}}}}{{n\left( {n - 1} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {x^n},{\text{ }}x = 1,{\text{ }}n \geqslant 2 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {{x^n}} \right] \cr
& y'\left( x \right) = n{x^{n - 1}} \cr
& {\text{Evaluate at }}x = 1 \cr
& y'\left( 1 \right) = n{\left( 1 \right)^{n - 1}} = n \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {n{x^{n - 1}}} \right] \cr
& y''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}} \cr
& {\text{Evaluate at }}x = 1 \cr
& y''\left( 1 \right) = n\left( {n - 1} \right){\left( 1 \right)^{n - 2}} \cr
& y''\left( 1 \right) = n\left( {n - 1} \right) \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 1 \cr
& K = \frac{{\left| {n\left( {n - 1} \right)} \right|}}{{{{\left( {1 + {{\left[ n \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{{n\left( {n - 1} \right)}}{{{{\left( {1 + {n^2}} \right)}^{3/2}}}} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{{{\left( {1 + {n^2}} \right)}^{3/2}}}}{{n\left( {n - 1} \right)}} \cr} $$
