Let $A$ be any matrix and $A=SJS^{-1}$ its Jordan decomposition. Then, by the definition of the exponential function for matrices, we obtain:
$$ \exp(At)=\exp(SJS^{-1}t) = \sum_{i=0}^{\infty} \frac{(SJS^{-1})^i t^i}{i!} = S\sum_{i=0}^{\infty} \frac{J^it^i}{i!}S^{-1}=S\exp(Jt)S^{-1}$$
since the $S$ cancel in the powers. Now, given $J=\bigoplus_j J_j$ (decomposition of $J$ into Jordan blocks), you have, since the Jordan blocks commute:
$$ \exp(Jt)=\bigoplus_j \exp{J_jt}$$
In particular, if $A$ is diagonalizable, then $\exp(J)$ is also diagonal and the diagonal entries are just $\exp(J_jt)$, with $J_j$ the eigenvalues. Otherwise, you need to calculate the exponential function of the Jordan blocks. This already tells you that there is no "constant c" the way you want to have it. The constant changes with the different Jordan blocks.
Your Jordan blocks will look like $J:=\lambda I + N$ with a diagonal part and an nilpotent part $N$. Since the two parts commute (check this!) we have:
$$\exp(Jt)=\exp(\lambda I t)\exp(Nt)=\exp(\lambda t)\exp{Nt}$$
which might help for calculations.
The linear operators $D$ and $N$ commute, which means that the exponential properties apply:
$$
e^{t(D+N)} = e^{tD}e^{tN} = e^{tN}e^{tD}.
$$
The exponential of a diagonal is easy enough to compute in any basis for which $D$ is diagonalized. Nilpotent matrices act like higher order differentials when plugging into a power series $F(z)$. For example, if $N$ is nilpotent of order $m$, then
$$
F(N)=\frac{F(0)}{0!}I+\frac{F'(0)}{1!}N+\frac{F''(0)}{2!}N^{2}+\cdots+\frac{F^{(m-1)}(0)}{(m-1)!}N^{m-1}.
$$
This breaks down nicely when you restrict to one Jordan block. This is because $D=\lambda I$ in that case, which gives $e^{tD}=e^{t\lambda}I$. For example,
$$
\exp\left\{t
\begin{pmatrix}
\lambda & 1 & 0 & 0 \\
0 & \lambda & 1 & 0 \\
0 & 0 & \lambda & 1 \\
0 & 0 & 0 & \lambda
\end{pmatrix}\right\} =
e^{t\lambda}\begin{pmatrix}
1 & \frac{t}{1!} & \frac{t^{2}}{2!} & \frac{t^{3}}{3!} \\
0 & 1 & \frac{t}{1!} & \frac{t^{2}}{2!} \\
0 & 0 & 1 & \frac{t}{1!} \\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
This follows because
$$
N = \begin{pmatrix}0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0\end{pmatrix},
N^{2} = \begin{pmatrix}0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{pmatrix},
N^{3} = \begin{pmatrix}0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{pmatrix}.
$$
By the way, this is all closely related to differential operators. For example, if you want the solutions to $(\frac{d}{dx}-\lambda)^{4}f = 0$, the solutions are combinations of
$$
e^{t\lambda},\; e^{t\lambda}\frac{t}{1!},\; e^{t\lambda}\frac{t^{2}}{2!},\; e^{t\lambda}\frac{t^{3}}{3!}.
$$
And $(\frac{d}{dx}-\lambda)$ maps the 4th one to the 3rd, the 3rd to the 2nd, etc., and finally maps the first to $0$.
Best Answer
With
$A = SJS^{-1}, \tag 1$
we have
$Se^{At}S^{-1} = S \left (\displaystyle \sum_0^\infty \dfrac{(At)^n}{n!} \right ) S^{-1}$ $= \displaystyle \sum_0^\infty \dfrac{(SAS^{-1}t)^n}{n!} = \sum_0^\infty \dfrac{(Jt)^n}{n!} = e^{Jt}; \tag 2$
that is, $e^{At}$ is similar to $e^{Jt}$; thus their characteristic polynomials are the same, since
$\det ( e^{Jt} - \lambda I) = \det (Se^{At}S^{-1} - \lambda SS^{-1}) = \det (S(e^{At} - \lambda I)S^{-1})$ $= \det (S) \det(e^{At} - \lambda I) \det (S^{-1}) = \det(e^{At} - \lambda I), \tag 3$
since
$\det(S) \det(S^{-1}) = \det (SS^{-1}) = \det(I) = 1; \tag 4$
since the characteristic polynomials are the same, the eigenvalues, being the roots of said polynomials, are also the same, that is
$\text{eig}(e^{At}) = \text{eig}(e^{Jt}), \tag 5$
$OE\Delta$.