Find the Cauchy product of $e^x$ and $e^{-x}$.

Question

I was able to get to:

$c_n$ = $$\sum_{k=0}^{n} \binom{n}{k} \frac{x^k (-x)^{n-k}}{n!}$$ but I am stuck as to how to get past this. Is there a property of the series that lets me pull the negative sign out? I am trying to show that all coefficients aside from $k = 0$ will give me $0$. We are given the hint of Newton's binomial formula, and I presume I will want to have my second term in that be zero, but I am unsure how to show that the numerator of the series is always zero.

Answer 1

Hint:$$\sum_{k=0}^{n}\binom{n}{k}\frac{x^k(-x)^{n-k}}{n!}=x^n\sum_{k=0}^{n}\frac{(-1)^k}{k!(n-k)!}$$

Answer 2

Note that $x^k(-x)^{n-k}=(-1)^{n-k}x^n$ and that$$\sum_{k=0}^n\binom nk(-1)^k=\bigl(1+(-1)\bigr)^n=0$$if $n>0$.