Let $\lambda\in \sigma(A).$ Then, $e^{\lambda t}\in \sigma(e^{tA})$ where $\sigma(A)$ is set of all eigenvalues of matrix $A$ Let $\lambda\in \sigma(A).$ Then, $e^{\lambda t}\in \sigma(e^{tA})$ where $\sigma(A)$ is set of all eigenvalues of matrix $A$.
MY TRIAL
Let $\lambda\in \sigma(A).$ Then, $\exists:t\neq 0$ s.t. 
\begin{align}Ax=\lambda x  \end{align}
\begin{align}\sum^{\infty}_{n=0}\frac{A^n}{n!}=\sum^{\infty}_{n=0} \frac{ \lambda^n}{n!} \end{align}
\begin{align}e^{A}=e^{\lambda } \end{align}
So,
\begin{align}e^{At}=\sum^{\infty}_{n=0}\frac{A^n t^n}{n!}=\sum^{\infty}_{n=0}\frac{\lambda^n t^n}{n!}=e^{\lambda t}\end{align}
and we're done!
I'm skeptical about my proof. Can someone please, check for me? If the proof is wrong, alternative proofs will be highly regarded. Thanks

Your idea is right but your proof is muddled (due to mixing up the scalar $t$ with the eigenvector of $A$ corresponding to $\lambda$).
Suppose $v$ is an eigenvector with $Av = \lambda v$. Then
$$e^{At}v = \sum_{i=0}^{\infty} \frac{(At)^i}{i!}v = \sum_{i=0}^{\infty} \frac{t^i}{i!} (A^i v) = \sum_{i=0}^{\infty} \frac{(\lambda t)^i}{i!} v = e^{\lambda t}v.$$
Therefore $v$ is an eigenvector of $e^{At}$ with eigenvalue $e^{\lambda t}$. The one additional detail that you may want to check (depending on the level of rigor expected) is that the sums in the above calculation are well-defined (they are).
