Exponential Integrals: the derivative of the exponential integral $E_n(x)$

Question

The exponential integral is defined as: $$E_n=x^{n-1} \int_x^\infty{\frac{e^{-t}}{t^n}}dt,$$ and its derivative with respect to $x$ should be: $$\frac{dE_n}{dx}=-E_{n-1}.$$
However, when I tried to prove this relation I got stuck at: $$\frac{dE_n}{dx}=(n-1)x^{n-2}\int_x^\infty{\frac{e^{-t}}{t^n}}dt-\frac{1}{xe^x},$$ and I don't know how this should equal $-E_{n-1}.$

Answer 1

Note that: $$\int_{x}^{\infty}\frac{e^{-t}}{t^{n}}dt = \frac{e^{-t}t^{-n+1}}{-n+1}\bigg{|}_{x}^{\infty}+\frac{1}{n-1}\int_{x}^{\infty}\frac{e^{-t}}{t^{n-1}}dt.$$ Thus, $$(n-1)x^{n-2}\int_{x}^{\infty}\frac{e^{-t}}{t^{n}}dt = \frac{1}{xe^{x}}-x^{n-2}\int_{x}^{\infty}\frac{e^{-t}}{t^{n-1}}dt = \frac{1}{xe^{x}}-E_{n-1}.$$