Duplicate - Proof by Ordinary Induction: $a^n-b^n \leq na^{n-1}(a-b)$ This question has already been asked:

Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$

However has not been answered properly. (Even thought the OP checked an answer)

The answers provided are direct proofs and not induction proofs.

Question: Suppose $a$ and $b$ are real numbers with $0 &lt; b &lt; a$. Prove that if $n$ is a positive integer, then:

$$a^n-b^n \leq na^{n-1}(a-b)$$

Anybody has ideas?
We do the induction step. Suppose that for a certain $k$ we have $a^k-b^k\le ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}\le (k+1)a^k(b-a)$.

We have
$$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).\tag{1}$$
Note that by the induction hypothesis $b(a^k-b^k)\le a(a^k-b^k)\le ka^k(a-b)$. So the sum on the right-hand side of (1) is $\le (k+1)a^k(a-b)$.
The proof proceeds by induction. Here we will need to make use of two base
cases, since it turns out that n = 1 needs to be treated separately.

(1) First base case (n = 1): here a^1 −b^1 ≤ 1*a^0(a−b) = a − b,


so the statement holds.

(2) Second base case (n = 2):
a^2 − b^2 = (a + b)(a − b)
&lt; 2a(a − b)    since a &gt; b
= na^(n−1)(a − b), for n = 2

(3) Inductive step: Now suppose that a^n − b^n &lt; na^(n−1)(a − b), where 0 &lt; 
b &lt; a. Then
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
&lt; a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
&lt; a^n(a − b) + ana^(n−1)(a − b) since b &lt; a
= (n + 1)a^n(a − b)
so that a^(n+1) − b^(n+1) &lt; (n + 1)a^n(a − b) for n &gt; 2.