Answer
The interval of convergence is $(-\infty,\infty)$.
Work Step by Step
We apply the ratio test
$$
\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ 8^{n+1}x^{n+1}/(n+1)! }{ 8^{n}x^{n}/n! }|=8|x|\lim _{n \rightarrow \infty} \frac{1}{(n+1)}=8|x|\times0
$$
Hence, the series $\Sigma_{n=0}^{\infty} 8^{n}x^{n}/n!$ converges if and only if $|x|\times0=0\lt1$ which is true for any value of $x$. That is, the interval of convergence is $(-\infty,\infty)$.
