If all the eigen values are distinct then all the eigen vectors are distinct. Indeed let $\lambda_1\neq\lambda_2$ two eigen values, let $x_1$ an eigenvector of $\lambda_1$, let's show that $x_1$ can't be an eigenvector of $\lambda_2$.
If $x_1$ is also an eigenvector of $\lambda_2$, then : $Ax_1=\lambda_2 x_1\Rightarrow \lambda_1 x_1=\lambda_2 x_1\Rightarrow x_1(\lambda_1-\lambda_2)=0\Rightarrow \lambda_1-\lambda_2=0$. Because $x_1 \neq0$ because it is an eigenvector. So $\lambda_1=\lambda_2$, it is a contradiction.
So $A$ has $n$ different eigenvectors and you can conclude withe the result you mentionned.