Laguerre Polynomials

Python 2, 53 bytes

f=lambda n,x:n<1or((2*n-1-x)*f(n-1,x)-~-n*f(n-2,x))/n

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Wolfram Language (Mathematica), 9 bytes

LaguerreL

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Jelly, 11 bytes

cŻ÷Ż!$ƲṚḅN}

A dyadic Link accepting \$n\$ on the left and \$x\$ on the right which yields \$L_n(x)\$.

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How?

This makes the observation that
\$L_n(x)=\sum\limits_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k!}x^k=\sum\limits_{k=0}^{n}{(-x)^k}\frac{n\choose k}{k!}\$
which is the evaluation of a base \$-x\$ number with n+1 digits of the form \$\frac{n\choose k}{k!}\$.

cŻ÷Ż!$ƲṚḅN} - Link: n, x
      Ʋ     - last four links as a monad - f(n):
 Ż          -   zero-range (n) -> [0, 1, 2, ..., n]
c           -   (n) binomial (that) -> [nC0, nC1, nC2, ..., nCn]
     $      -   last two links as a monad - g(n):
   Ż        -     zero-range (n) -> [0, 1, 2, ..., n]
    !       -     factorial (that) -> [0!, 1!, 2!, ..., n!]
  ÷         -   division -> [nC0÷0!, nC1÷1!, nC2÷2!, ..., nCn÷n!]
       Ṛ    - reverse -> [nCn÷n!, ..., nC2÷2!, nC1÷1!, nC0÷0!]
          } - use the chain's right argument for:
         N  -   negate -> -x
        ḅ   - convert from base (-x) -> -xⁿnCn÷n!+...+-x²nC2÷2!+-x¹nC1÷1!+-x°nC0÷0!