Can we find $c, c'$ such that $ c_1 e^{-c_2 n^{c_3}}\leq e^{-cn^{c'}}?$ Given some constants $c_1, c_2, c_3$, can we find $c, c'$ such that 
$$ c_1 e^{-c_2 n^{c_3}}\leq  e^{-cn^{c'}}?$$
 A: I assume that you are working with the real numbers. First note that if $c_1 \leq 0$, the given inequality clearly holds for all values of $c,c'$. Thus we restrict ourselves to the cases where $c_1 >0$.
Also, when $n=0$, L.H.S. = $c_1$ and R.H.S. = $1$, so we simply have to choose $c_1 \leq 1$. So we discuss below the cases where $n \neq 0$.
Now, $c_1 e^{-c_2 n^{c_3}}\leq  e^{-cn^{c'}} \Rightarrow c_1e^{-c_2n^{c_3}+cn^{c'}}\leq 1 \Rightarrow e^{-c_2n^{c_3}+cn^{c'}} \leq \frac{1}{c_1} $
Hence $-c_2n^{c_3}+cn^{c'} \leq -\ln(c_1) \Rightarrow cn^{c'}\leq c_2n^{c_3}-\ln(c_1)$. This is because taking the natural logarithm is an order-preserving operation.
If $0<c_1\leq1$, such that $\ln{c_1}\leq 0 \Rightarrow -\ln{c_1}\geq 0 $, simply set $c=c_2,c'=c_3$ and you are done.
Otherwise, if $c_1>1$, let  $c'=c_3$, so $cn^{c_3}\leq c_2n^{c_3}-\ln(c_1) \Rightarrow c \leq c_2-\frac{\ln(c_1)}{n^{c_3}}$ if $n^{c_3} >0$ (i.e. just choose any value of $c$ satisfying this inequality since the values of $c_1,c_2,c_3$ are already known), and $c \geq c_2-\frac{\ln(c_1)}{n^{c_3}}$ if $n^{c_3} < 0$. This is, of course, assuming that $n^{c_3}$ is well-defined.
Thus, we may conclude that there always exist such numbers $c$ and $c'$. (Of course, this is assuming your exponentiation operation $n^{c_3}$ is defined).
