TITLE: $n$th derivative of $e^{ax^m}$
QUESTION [7 upvotes]: I am seeking an explicit expression for the $n$th derivative of $e^{ax^m}$.

To achieve this I consider the $n$th coefficient of the Taylor series with $t\rightarrow 0$.
\begin{align}
    e^{a(x+t)^m} &= \sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a(x+t)^m}\frac{t^n}{n!}\\
    &=\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\left[\sum\limits_{k=0}^{m}\binom{m}{k}x^kt^{m-k}\right]}\frac{t^n}{n!}\\
    &=e^{ax^m}\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\sum\limits_{k=0}^{m-1}\binom{m-1}{k}x^kt^{m-k-1}}\frac{t^n}{n!}
\end{align}
But we can also consider the left hand side as
\begin{align}
e^{a(x+t)^m} &= \sum_{k=0}^\infty\frac{(a(x+t)^m)^k}{k!}\\
&= \sum_{k=0}^\infty\frac{a^k}{k!}\sum_{j=0}^{mk}\binom{mk}{j}x^{j}t^{mk-j}\\
&= \sum_{k=0}^\infty\frac{a^k}{k!}x^{mk}\sum_{j=0}^{mk-1}\binom{mk-1}{j}x^{j}t^{mk-1-j}\\
&= \sum_{k=0}^\infty\frac{(ax^m)^k}{k!}\sum_{j=0}^{mk-1}\frac{(mk-1)!}{j!(mk-1-j)!}x^{j}t^{mk-1-j}
\end{align}
Although I can't really take it further. My next step would be to extract the $n$th coefficient and, upon comparison to the first formula, obtain an expression for the $n$th derivative. When $m=2$ we obtain the Hermite polynomials. Related to
$n$th derivative of $e^{1/x}$
Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$
and links therein. With thanks to BillyJoe from the comments, Wolfram gives
\begin{equation}
\frac{d^n}{dx^n}e^{ax^m}=e^{ax^m}x^{-n}\sum^n_{k=0}\sum^k_{j=0}\frac{(-1)^j(ax^m)^k(1-jm+km-n)_n}{j!(k-j)!}
\end{equation}
I have also seen that this is equivalent to n exponent rial Riordan array, $B_m=[1,(1+x)^m+1]$.
\begin{equation}
\frac{d^n}{dx^n}e^{ax^m}=\left( \sum_{k\in \mathbb Z}b_{n,k;m}(ax)^{km-n}\right)e^{ax^m}
\end{equation}

REPLY [2 votes]: Hint:
Experimentally, the expression is of the form
$$(e^{x^m})^{(n)}=P_{mn}(x^m)x^{(-n)\bmod m}e^{x^m}$$ where $P_{mn}$ is a polynomial of degree $n-\left\lfloor\dfrac nm\right\rfloor$. You can arrange the coefficients in a triangular table, with a recurrence relation between the rows.
$$P_{m.n+1}x^{(-n-1)\bmod m}=\\mx^{m-1}(P'_{mn}(x^m)+P_{mn}(x^m))x^{(-n)\bmod m}+((-n)\bmod m)P_{mn}(x^m)x^{(-n)\bmod m-1}$$
which simplifies differently when $m|n$ or not.