Showing $|\exp(tA)| \leq K$ if all eigenvalues have real part negative or zero (and if zero real part, simple eigenvalue) I wish to show that if all eigenvalues have real part negative or zero and if those eigenvalues with zero real part are simple, there exists a constant $K>0$ such that $|\exp(tA)| \leq K$, $(0<t<\infty)$, and hence every solution of $y' = Ay$ is bounded on $(0 \leq t < \infty)$. 
We can write our matrix $A$ as $A = PBP^{-1}$ and since all eigenvalues are simple, $B$ is diagonal so $A^n = PB^nP^{-1}$. If $B$ is nilpotent then our sum has finite number of terms and stops at the $n-1$ term. 
EDIT:
 A: In this answer, I will focus on showing that $\Vert e^{At} \Vert$ is bounded where $\Vert \cdot \Vert$ is the standard operator norm,
$\Vert A \Vert = \sup \{\Vert Ax \Vert, \; \Vert x \Vert = 1 \} = \inf \{C \mid \Vert Ax \Vert \le C \Vert x \Vert, \; \forall x \}, \tag 0$
where $\Vert x \Vert$ is the standard Hermitian norm for vectors $x \in \Bbb C^n$, that is, $\Vert x \Vert$ is derived from the Hermitian inner product
$\langle y, z \rangle = \displaystyle \sum_1^n \bar y_i z_i, y = (y_1, y_2, \ldots, y_n)^T \in \Bbb C^n, \; \text{etc,} \tag{0.1}$ 
by
$\Vert x \Vert^2 = \langle x, x \rangle =  \displaystyle \sum_1^n \bar x_i x_i. \tag{0.2}$
We resort to extending the normal real Euclidean inner product on $\Bbb R^n$ to the Heritian product (0.1) to facilitate addressing and handling situations in which some of the eigenvalues of $A$ are non-real complex numbers.
We want to show that
$\forall t \in \Bbb R, \; 0 \le t < \infty, \Vert e^{At} \Vert < K \tag 1$
for some $0 < K \in \Bbb R$.
We know that $A$ may be cast into Jordan canonical form by a similarity transformation
$A \to PAP^{-1} \tag 2$
for some non-singular matrix $P$; we thus first deal with the case of $A$ a Jordan matrix; indeed, we will first address the situation when $A$ is a single Jordan block, that is, a matrix of the form 
$A = \lambda I_m + N, \tag 3$
here $I_m$ is the $m \times m$ identity matrix, $\lambda$ is an eigenvalue of $A$ and $N$ is the $m \times m$ nilpotent matrix consisting of $m - 1$ $1$s on the superdiagonal and zeroes everywhere else.  Since $I_m$ commutes with $N_m$, that is,
$I_m N_m = N_m = N_m I_m, \tag 4$
it follows that
$e^{At} = e^{(\lambda I_m + N_m)t} = e^{\lambda I_m t + N_m t} = e^{\lambda I_m t} e^{N_m t}; \tag 5$
now it is both well-known and easy to see from the matrix power series of $\exp$ that
$e^{\lambda I_m t} = \displaystyle \sum_0^\infty \dfrac{(\lambda I_m t)^k}{k!} = \sum_0^\infty \dfrac{\lambda^k t^k I_m }{k!} = \sum_0^\infty \dfrac{\lambda^k t^k}{k!} I_m = e^{\lambda t} I_m; \tag 6$
also, since the nilpotent matrix $N$ satisfies
$N^m = 0, \tag 7$
we have
$e^{N_m t} = \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!}, \tag 8$
which is an $m \times m$ matrix whose entries are polynomials in $t$ of degree at most $m - 1$; it follows from (5), (6) and (8) that $e^{At}$ takes the form
$e^{At} =  e^{\lambda t} I_m \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} = e^{\lambda t}  \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!}, \tag 9$
and if
$\lambda = \sigma + i \omega, \; \sigma < 0, \tag{10}$
we may further decompose $e^{At}$ as
$e^{At} = e^{i\omega t} e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!}, \tag{11}$
whence
$\Vert e^{At} \Vert = \left \Vert  e^{i\omega t} e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} \right \Vert = \vert e^{i\omega t} \vert \left \Vert e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} \right \Vert = \left \Vert e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} \right \Vert. \tag{12}$
Now it is well-known that for $\sigma < 0$ the expression on the right of (12), being dominated by the exponential $e^{\sigma t}$, eventually decreases to $0$ as $t \to \infty$:
$\displaystyle \lim_{t \to \infty} \Vert e^{At} \Vert = \lim_{t \to \infty} \left \Vert e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} \right \Vert = 0; \tag{13}$
since 
$\displaystyle  \Vert e^{At} \Vert = \left \Vert e^{\sigma t} \displaystyle \sum_0^{m - 1} \dfrac{N_m^k t^k}{k!} \right \Vert. \tag{14}$
is continuous as a function of $t$, it is bounded on any compact interval $[0, \tau]$; by choosing $\tau$ sufficiently large we may, in the light of the limit (13), assume $\Vert e^{At} \Vert < \epsilon$, $0 < \epsilon \in \Bbb R$, for $t \ge \tau$; therefore (14) is bounded by some $0 < K \in \Bbb R$ for all $t \in [0, \infty)$.  We note that in the case $\sigma = 0$, we have $m = 1$ by our hypotheses on the matrix $A$, and the Jordan block reduces to $e^{i\omega t}$, which is manifestly bounded; indeed, as long as the block size is $1$, the block reduces to $e^{(\sigma + i \omega)t}$ with $\sigma \le 0$ and is thus easily seen to be bounded in norm.
The preceding discussion shows that $e^{At}$ is bounded for any Jordan block (3) of $A$ as long as $\sigma = \Re(\lambda) \le 0$; if $A$ is comprised of multiple Jordan blocks, then there is a bound for each and hence, since the number of Jordan blocks is finite, a bound over all the Jordan blocks of $A$ is obtained by taking the greatest such bound.  
The preceding discussion covers the situation when $A$ is in Jordan form; it result may be extended to any $A$ by inverting the similarity transformation (2), an operation which preserves the boundedness of $\Vert e^{At} \Vert$ (though it may alter a particular bound) since $P$ does not depend upon $t$.  We conclude that, in all specified/required cases, there is some $K > 0$ with
$\Vert e^{At} \Vert < K, \; t \in [0, \infty). \tag{15}$
And we are done.
