We have, ${e^{ – 1}} = 1 + ( – 1) + \frac{{{{( – 1)}^2}}}{{2!}} + \frac{{{{( – 1)}^3}}}{{3!}} + \frac{{{{( – 1)}^4}}}{{4!}} + ……….$

The given integral = $\int {{{({e^{ – 1}})}^x}dx} = \int {{e^{ – x}}dx} = – {e^{ – x}} + C$