I have a minor result which I'm sure has come up somewhere before but I can't seem to find it.

Consider a confluent hypergeometric function of the form $$\newcommand{\ff}{{}_1F_1} \ff(b+k;b;z)\textrm{, for }k\in\mathbb{N}.$$ Numerical tests suggest that this is always a polynomial of degree $k$ multiplied by an exponential. One can prove this in a dull fashion by using $\ff(b;b;z)=e^z$ and then applying recurrence relations, but I found a cleaner way using the series definition,

$$ \ff(b+k;b;z) =\sum_{n=0}^\infty\frac{(b+k)_n}{(b)_n}\frac{z^n}{n!} $$

where $(b)_k=b(b+1)\cdots(b+k-1)$ is the Pochhammer symbol. By exploiting the identities $$ \frac{(b+k)_n}{(b)_n}=\frac{\Gamma(b+k+n)\Gamma(b)}{\Gamma(b+k)\Gamma(b+n)}=\frac{(b+n)_k}{(b)_k} \qquad\textrm{and}\qquad nz^n=z\frac{d}{dz}z^n, $$ one can easily prove that $$ \ff(b+k;b;z)=\frac{\left(b+z\frac{d}{dz}\right)_k}{(b)_k}e^z,$$

by being somewhat liberal with the meaning of the Pochhammer symbol. This is clearly the desired polynomial-times-exponential, and provides an explicit expression for the polynomial that looks kind of like a Rodrigues formula.

Even better, if you put this together with Kummer's first transformation, $$\ff\left(a;b;z\right)=e^{z}\ff\left(b-a;b;-z\right),$$ and the expression for Laguerre polynomials in terms of hypergeometric functions, $L^{(\alpha)}_{n}\left(x\right)=\frac{\left(\alpha+1\right)_{n}}{n!}\ff\left(-n,\alpha+1,x\right)$, you get an analogous result for Laguerre polynomials,

$$ L^{(b-1)}_{k}\left(x\right)=\frac{1+k/b}{k!}e^z\left(b+z\frac{d}{dz}\right)_ke^{-z}. $$

Are these results familiar to anyone? Do they fit inside a larger framework? They are not the best thing since sliced bread but they do have a nice simplicity to them, and particularly I would like to cite the appropriate reference if they have appeared before.