Question 115186
{{{(n+1)^k-n^k}}} Start with the given expression



*[Tex \LARGE (n^k+kn^{k-1}+\frac{k(k-1)}{2}n^{k-2}*1^2\ldots 1^k) -n^k] Expand {{{(n+1)^k}}} using the Binomial Theorem


*[Tex \LARGE (kn^{k-1}+\frac{k(k-1)}{2}n^{k-2}*1^2\ldots 1^k)+ n^k-n^k] Group like terms


Notice the {{{n^k}}} terms cancel out since {{{n^k-n^k=0}}}. So we're left with 


*[Tex \LARGE kn^{k-1}+\frac{k(k-1)}{2}n^{n-2}*1^2\ldots 1^k] 


Now the term with the highest exponent value is {{{kn^(k-1)}}}. So this means that the degree of {{{(n+1)^k-n^k}}} is {{{k-1}}}