[Math] Prove that there is no term independent of $x$ in the binomial expansion of $\left(x-\frac 1x\right)^{19}$

Question

I am dealing with a fairly simple question but I'm struggling a bit to come up with a formal demonstration on why the binomial expansion of $\left(x-\frac 1x\right)^{19}$ doesn't have a term "independent of $x$".

I am looking for a fairly straightforward analytical explanation involving the properties of the sigma operator for this particular case. Anything but "simply develop the entirety of the binomial and look for "x-less" terms.

Thank you so, so much in advance. Cheers!

Answer 1

Note that $\left(x-\frac 1x\right)^{19}=x^{-19}(x^2-1)^{19},$ so the question is the same as asking whether $(x^2-1)^{19}$ has a term $x^{19}$. But this polynomial is even, so it has no odd terms at all!