How do you find the general expression for the k^{th} derivative of an exponetial function with a function in the exponent? I'm looking for a general expression for the function $\frac{\delta^k}{\delta \mu^k}[e^{n\mu + \mu^2}]_{\mu=0}$
I was thinking I could use the taylor expansion coefficients, but the function in the exponent is throwing me off.
Edit 1:
Here's my newest attempt
Using the Taylor series coefficients we have
            $e^{n\mu + \mu^2} = \sum_{j=0}^{\infty} \frac{(n+\mu)^j}{j!}\mu^j$
So now we can evaluate our expression above
            $\frac{\delta^k}{\delta\mu^k}\left [ e^{n\mu + \mu^2} \right]_{\mu=0}=(n+\mu)^k|_{\mu=0}=n^k$
 A: To understand what is going on, write out a few terms of each series:  $$e^{n\mu+\mu^2} = \left( \frac{(n\mu)^0}{0!} + \frac{(n\mu)^1}{1!} + \frac{(n\mu)^2}{2!} + \cdots \right)\left( \frac{\mu^0}{0!} + \frac{\mu^2}{1!} + \frac{\mu^4}{2!} + \cdots \right).$$  If you were to expand and collect like exponents in $\mu$, then you could see how this would give you the coefficient of $\mu^j$, $j = 0, 1, 2, \ldots$.  This suggests looking for the set of all nonnegative integers $k, m$ such that $j = k + 2m$.  If $j$ is odd, then $k$ must be odd; if $j$ is even, then $k$ is also even.  So this is what LutzL is referring to by "distinguish the even and odd case."
For example, if $j = 5$, then we could get this through the choices $$(k,m) = (1,2), (3,1), (5,0)$$ only.  If $j = 8$, then the choices are $$(k,m) = (0,4), (2,3), (4,2), (6,1), (8,0).$$  So if we let $$e^{n\mu + \mu^2} = \sum_{j=0}^\infty a_j \frac{\mu^j}{j!} = \sum_{i=0}^\infty a_{2i} \frac{\mu^{2i}}{(2i)!} + a_{2i+1} \frac{\mu^{2i+1}}{(2i+1)!},$$ then $$a_{2i} \frac{\mu^{2i}}{(2i)!} = \sum_{t=0}^i \frac{(n \mu)^{2t}}{(2t)!} \frac{\mu^{2(i-t)}}{(i-t)!} = \mu^{2i} \sum_{t=0}^i \frac{n^{2t}}{(2t)! (i-t)!}.$$  Notice how the $\mu^{2i}$ term is constant with respect to the summation and therefore factors out, which confirms the correctness of the internal sum.  So we have $$a_{2i} = (2i)! \sum_{t=0}^i \frac{n^{2t}}{(2t)!(i-t)!} = \frac{d^{2i}}{d\mu^{2i}}\left[ e^{n\mu + \mu^2} \right]_{\mu = 0}.$$  A similar approach applies to the odd derivatives, $j = 2i+1$.
A: $$
e^{nμ+μ^2}=e^{nμ}·e^{μ^2}=\sum_{m=0}^\infty \frac{(nμ)^m}{m!}·\sum_{j=0}^\infty\frac{μ^{2j}}{j!}
$$
Now order by exponents. It may be useful to distinguish the even and odd case.

The coefficient of degree $k$ is composed of all products of terms with $k=m+2j$. This starts with $j=0$ and $k=m$ and decreases $m$ in steps of 2. Or simply $m=k-2j$.
$$
c_k=\sum_{j=0}^{\lfloor k/2\rfloor}\frac{n^{k-2j}}{(k-2j)!}·\frac1{j!}
$$
