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Let x,yāFx,y\in\mathbb{F}x,yāF then (x+y)n=āk=0n(nk)xnākyk=āk=0n(nk)xkynāk(x+y)^{n}=\sum\limits_{k=0}^{n} {n\choose k} x^{n-k}y^{k}=\sum\limits_{k=0}^{n}{n\choose k}x^{k}y^{n-k}(x+y)n=k=0ānā(knā)xnākyk=k=0ānā(knā)xkynāk