$n$th derivative of $e^{1/x}$ I am trying to find the $n$'th derivative of $f(x)=e^{1/x}$. When looking at the first few derivatives I noticed a pattern and eventually found the following formula
$$\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}$$
I tested it for the first $20$ derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don't want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step.
Here is what I tried for the induction (incomplete, maybe incorrect)
$\begin{align*}
\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)&=\frac{\mathrm d}{\mathrm dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}\\
&=(-1)^n e^{1/x} \cdot \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1}\right)-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}\\
&=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})\\
&=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k}
\end{align*}$
I don't know how to get on from here.

Peter's and Mike's answers have clearly settled this question; I'll just explain the OP's mention of "Mathematica says that it is some hypergeometric distribution". More specifically, one wonders how Mathematica might have arrived at the Kummer confluent hypergeometric function ${}_1 F_1\left({{a}\atop{b}}\mid x\right)$.
We start with the transformed Maclaurin series:
$$\exp\left(\frac1{x}\right)=1+\sum_{k=1}^\infty \frac1{x^k k!}$$
and then differentiate the series termwise, using the formula
$$\frac{\mathrm d^n}{\mathrm dx^n}x^{-k}=\frac{(-1)^n (n+k-1)!}{(k-1)! x^{k+n}}$$
to yield
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=1}^\infty \frac{(n+k-1)!}{k!(k-1)! x^k}$$
which can be reindexed like so:
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=0}^\infty \frac{(n+k)!}{(k+1)!k! x^{k+1}}$$
and rewritten in terms of Pochhammer symbols to yield
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n}{x^n}\sum_{k=0}^\infty \frac{n!(n+1)_k}{(2)_k k! x^{k+1}}$$
or, expressed hypergeometrically,
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}{}_1 F_1\left({{n+1}\atop{2}}\mid \frac1{x}\right)$$
It's not immediately obvious that this one can be expressed in terms of $\exp(1/x)$ multiplied by a finite sum, but we can then apply the Kummer transformation to the hypergeometric function like so:
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right){}_1 F_1\left({{1-n}\atop{2}}\mid -\frac1{x}\right)$$
and find that the hypergeometric function has a nonpositive numerator parameter $1-n$ for $n > 0$, which means it can be expressed as a terminating series (since the Pochhammer symbol $(1-n)_k$ is $0$ for $k \geq n$):
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right)\sum_{k=0}^{n-1}\frac{(-1)^k (1-n)_k}{k!(k+1)!x^k}$$
Rewriting the Pochhammer symbol in terms of factorials, we get
$$\frac{\mathrm d^n}{\mathrm dx^n}\exp\left(\frac1{x}\right)=\frac{(-1)^n n!}{x^{n+1}}\exp\left(\frac1{x}\right)\sum_{k=0}^{n-1}\frac{(n-1)!}{(n-k-1)!k!(k+1)!x^k}$$
which is equivalent to the formula you obtained.
My point here, I guess, is that hypergeometric functions are nothing to be uncomfortable with; if it helps, just consider them as shorthand for certain (in)finite sums.
