Answer
The interval of convergence is $x=0$.
Work Step by Step
Let ${a_n} = {n^n}{x^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{{\left( {n + 1} \right)}^{n + 1}}{x^{n + 1}}}}{{{n^n}{x^n}}}} \right| = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\left( {n + 1} \right){{\left( {\dfrac{{n + 1}}{n}} \right)}^n}} \right|$
$ = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\left( {n + 1} \right){{\left( {1 + \dfrac{1}{n}} \right)}^n}} \right| = \infty $
We have $\rho = \infty $ for all $x$. However, at $x=0$ the series becomes $0$, so converges. Hence, $\mathop \sum \limits_{n = 0}^\infty {n^n}{x^n}$ diverges for all $x \ne 0$.
Hence, the interval of convergence is $x=0$.
