#### Answer

The function $f(x)$ is continuous on $\left [0 ,2\right]$ with $f(0)=0$ and $f(2)=2^n$
Therefore, by the Intermediate Value Theorem, there is a $c \in\left (0,2\right)$ such that $f(c)= c^n$. So, the equation $c=\sqrt [n] {c}$ has a solution $c$ in $\left[0,2\right]$.

#### Work Step by Step

We are given that $f(x)=x^n$
The function $f(x)$ is continuous on $\left [0 ,2\right]$ with $f(0)=0$ and $f(2)=2^n$
Therefore, by the Intermediate Value Theorem, there is a $c \in\left (0,2\right)$ such that $f(c)= c^n$. So, the equation $c=\sqrt [n] {c}$ has a solution $c$ in $\left[0,2\right]$.