$k(k^n - (k-1)^n) = k^{n+1}-(k-1)^{n+1}-(k-1)^n$

Question

In Feller's Intro to Probability Theory Vol. 1. there is a step I don't know how the author proceded.

You have the full source here

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In first place we have (you can ignore the $N^n$):

$$N^n p_k = k^n - (k-1)^n $$

Now we have:

$$E(X) = \sum_{k = 1}^Nkp_k$$ $$ = N^{-n}\sum_{k = 1}^N \{k (k^n - (k-1)^n)\} = N^{-n}\sum_{k = 1}^N \{k^{n+1}-(k-1)^{n+1}-(k-1)^n\}$$

Which is followed by the question:

$$k(k-1)^n \overset{?}{=} (k-1)^{n+1}+(k-1)^n$$

This algebra is the argument for an approximation he later does on the same page number, however I don't believe the last steps he made are correct.

Answer 1

$k(k-1)^n - (k-1)^n = (k-1)(k-1)^n = (k-1)^{n+1}$

and add $(k-1)^n$ to both sides.

Answer 2

Don't try expanding the expression out; try factoring the right-hand side: $$ (k - 1)^{n+1} = (k-1)^n(k-1) $$ so that $$ (k - 1)^{n + 1} + (k - 1)^{n} = (k-1)^{n}(k - 1) + (k - 1)^{n} = \left[(k - 1) + 1\right](k-1)^{n} = k(k-1)^{n}. $$