Formally using the inverse Mellin transform for x>0:
$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$
$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$
Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$
then
$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$
where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.
For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.
See also Rodrigues-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:
$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$
with $D=\tfrac{d}{dx}$, leading to
$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$
This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):
$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$
leading to
$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$
Edit (Sept. 2019):
Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":
$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$
$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$
where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.
Best Answer
Both assertions hold true, in fact the roots lie in the interval $$(0,k+(k-1)\sqrt{k}).$$
There are many books for references, one among which is
"Basic Hypergeometric Series", by George Gasper, Mizan Rahman, Encyclopedia of Mathematics and its applications.