= Solution
Write a second-order <linear ordinary differential equation> in monic form $y''+P(x)y'+Q(x)y=0$. At a finite point $x_0$, an <ordinary point> has analytic $P,Q$. A singular point is a <regular singular point> if $(x-x_0)P$ and $(x-x_0)^2Q$ extend analytically there. A singular point failing this condition is an <irregular singular point>. The point at infinity is classified by the reciprocal coordinate $t=1/x$.
For the <Laguerre differential equation>, $P=1/x-1$ and $Q=\nu/x$. Thus $xP=1-x$ and $x^2Q=\nu x$ are analytic at zero, while $P$ itself is singular: zero is regular singular. Put $Y(t)=y(1/t)$. Since $y'=-t^2Y'$ and $y''=t^4Y''+2t^3Y'$, the transformed equation is
$$
Y''+\left(\frac1t+\frac1{t^2}\right)Y'+\frac\nu{t^3}Y=0.
$$
Here $tP_t=1+1/t$ is not analytic at zero, so \b[infinity is an irregular singular point], for every real $\nu$.
For the <Frobenius solution>, the lowest power $x^{\sigma-1}$ has coefficient $a_0\sigma^2$. Hence the <indicial equation> is $\sigma^2=0$, and equating the remaining powers gives
$$
(k+1+\sigma)^2a_{k+1}+(\nu-k-\sigma)a_k=0.
$$
With the repeated root $\sigma=0$, the <Frobenius series for the Laguerre differential equation> obeys
$$
\boxed{a_{k+1}=\frac{k-\nu}{(k+1)^2}a_k.}
$$
If the series does not terminate, its consecutive term ratio is asymptotic to $x/k$, so the <ratio test> gives infinite <radius of convergence>. It is an <entire function>, even though the differential equation has a singular coefficient at zero. The repeated indicial root yields only one analytic Frobenius branch; an independent solution generally has a logarithmic term there.
For $\nu=n\geq0$, the factor $k-n$ stops the recurrence at degree $n$. With $a_0=1$ its coefficients are $a_k=(-1)^k\binom nk/k!$, giving the <Laguerre polynomials>
$$
\boxed{L_0=1,\qquad L_1=1-x,\qquad L_2=1-2x+\frac{x^2}{2},\qquad L_3=1-3x+\frac{3x^2}{2}-\frac{x^3}{6}.}
$$
The same termination argument includes $n=0$, as required for $L_0$ despite the earlier reference to positive integers.
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