Can the ratio test be used to solve the following sequence?

You have that $$ \begin{gathered} \mathop {\lim }\limits_{k \to + \infty } \frac{{(k + 1)^{100} }} {{ek^{100} }} = \frac{1} {e}\mathop {\lim }\limits_{k \to + \infty } \left( {\frac{{k + 1}} {k}} \right)^{100} = \hfill \\ \hfill \\ = \frac{1} {e}\mathop {\lim }\limits_{k \to + \infty } \left( {\frac{{k + 1}} {k}} \right)^{100} = \frac{1} {e} \cdot 1^{100} = \frac{1} {e} < 1 \hfill \\ \end{gathered} $$ therefore your series is convergent

Your way is right, to conclude we simply have

$$\frac{(k+1)^{100}}{ek^{100}}=\frac1e \left(\frac{k+1}{k}\right)^{100}\to \frac1e \cdot 1^{100}=\frac1e <1$$

indeed

$$\frac{k+1}{k}=1+\frac1k \to 1 \implies \left(\frac{k+1}{k}\right)^{n} \to 1^n=1 \quad \forall n$$