By the continued fraction of the exponential integral, $$E_1(x)=\cfrac{e^{-x}}{x+\cfrac{1}{1+\cfrac{1}{x+\cfrac{2}{1+\cfrac{2}{\cfrac{\cdots}{x+\cfrac{n}{1+\cfrac{n}{x}}}}}}}}+R_n$$ where $R_n$ is a remainder.
I came up with the following formula. $$R_n=\frac{n+1}{-x L_{n}^{(1)}(-x)}\sum_{k=1}^{n+1}{{n}\choose{k-1}}(-1)^kE_{k+1}(x)$$ Or equivalently $$R_n=\frac{n+1}{-x L_{n}^{(1)}(-x)}\int_{0}^1{{e^{-x/t}}(1-t)^ndt}$$ where $L_{k}^{(\alpha)}(x)$ is the generalized Laguerre polynomial.
I figured this out a long time ago and now I can’t remember how I did it. Any help is appreciated.