I have a function $f(t)=\exp\left(\lambda\left(-1+\frac{1}{1-t}\right)\right)$. By taking the $n$'th derivative of $f$ and setting $t=0$, I will acquire a polynomial:
$$P(n) =\sum_{i=1}^{n} a_{n}(i) \lambda^{i}.$$
For example $P(2) =2*\lambda +\lambda^{2}$. Is there a general formula for acquiring $P(n)$?
Yes, Mathematica gives $$P(n) = n!\cdot L_{n}^{(-1)}(-\lambda),$$ where $L_{n}^{a}(x)$ denotes the generalised Laguerre polynomials. Using the explicit formula for these polynomials, we can also write $$P(n) = n!\sum_{k=1}^{n}\binom{n-1}{n-k}\frac{\lambda^k}{k!},$$ meaning that the coefficient of $\lambda^k$ in the polynomial $P(n)$ is $$\frac{n!}{k!}\binom{n-1}{n-k} = \frac{n!(n-1)!}{k!(k-1)!(n-k)!} = \frac{(n-1)!}{(k-1)!}\binom{n}{k}.$$