I did the following exercise : "Find Maclaurin series for $e^{-5x}$. "
My result was $e^{-5x} = (e^{-x})^5 = \left[\sum_{i=0}^{\infty}\frac{(-1)^nx^n}{n!} \right]^5$, which I verified was correct with Wolfram Alpha.
In the book they had the result in expanded form like so : $e^{-5x} = 1 - 5x + \frac{5^2x^2}{2!} - \frac{5^3x^3}{3!} + \frac{5^4x^4}{4!} - ...$
I'm pretty sure this is some kind of application of the Multinomial Theorem or something? I was just wondering, since there so obviously is a pattern here, what rule is applied here to find the first several terms of an infinite series raised to some power $k$?
Since $e^t=\sum_{i=0}^\infty \frac{t^i}{i!}$, just take $t=-5x$ to get $$e^{-5x}=\sum_{i=0}^\infty \frac{(-5x)^i}{i!}=\sum_{i=0}^\infty \frac{(-1)^i 5^i x^i}{i!} = 1-5x+\frac{5^2}{2!}x^2-\frac{5^3}{3!}x^3+\frac{5^4}{4!} x^4+\ldots$$