Monomials in terms of binomial coefficients

Question

Is there an explicit expression (or at least a recurrence relation) for the coefficients of a monomial $x^n$ in the basis of polynomials given by binomial coefficients $P_k(x) = \binom{x}{k}$, namely
$$
x^n = \sum_{k=0}^\infty a_{n,k} P_k(x) = \sum_{k=0}^\infty a_{n,k} \binom{x}{k},
$$
where the (generalized) binomial coefficient is given by $\binom{x}{k} = \frac{x(x-1)\dots(x-k+1)}{k!}$? I would be happy with a solution for $x\in\mathbb{N}$.

EDIT: The question arised when calculating sums of the form
$$
\sum_{l=1}^N P(l) \binom{N-l}{b},
$$
where $b\in\mathbb{N}$ and $P$ being a polynomial, which can easily be expressed as a sum of monomials. These sums arised from counting problems in quantum mechanics. Writing $P(l)=\sum_{n} b_n l^n$, and $l^n=\sum_k a_{n,k} \binom{l}{k}$, one can use an upper-index Vandermonde identity to calculate
$$
\sum_k a_{n,k} \sum_{l=1}^N \binom{l}{k} \binom{N-l}{b} = \sum_k a_{n,k} \binom{N+1}{k+b+1}.
$$

Answer 1

As noted in comments, we can express those coefficients in terms of Stirling numbers of the second kind.

Using the identity $\sum_{k=0}^{n}{n\brace k}(x)_k=x^n$, where $(x)_k$ is a falling factorial $$ (x)_k=x(x-1)\cdots(x-(k-1)), $$ we notice $(x)_k=\binom{x}{k}k!$, and so $x^n=\sum_{k=0}^{n}a_{n,k}\binom{x}{k}$ with $a_{n,k}=k!{n\brace k}$.

By the way, this generalizes to the following result:

For a polynomial $f(x)=\sum_{k=0}^{n}b_kx^k$, we can write $f(x)=\sum_{k=0}^{n}a_k \binom{x}{k}$ where $a_k=\sum_{i=k}^{n}k!{i\brace k}b_i$.