/sci/ - Science & Math

>>12516497
Another comfy one is
https://en.wikipedia.org/wiki/Wallpaper_group

>>12516355
Well we have that
[math](x^{n+1}-1)^n=(x-1)^n\left(\sum_{k=0}^n x^k\right)^n =_{n=2} \left(\sum_{k=0}^n x^k\right)^n[/math]
which further expands as a "relatively simple" multinomial expression (which have binomial coefficients with products of k! in the denominator, but all the x-terms are very basic)

and on the other hand the standard binomial expression
[math](x^n+1)^{n+1} = \sum_{k=0}^{n+1} {{n+1} \choose k} x^{n+(n+1-k)} [/math]
and you gotta e.g. argue that those binomial coefficients are each always smaller.

There's a chance that if you write down the multinomial expression explicitly, you can compare it to n+1 over k and find it's true on the nose.

There might also be a smarter way, I don't know..