Induction : Prove $a^k-b^k \le ka^{k-1}(a-b)$ I am doing revision for my discrete mathematics and I am having a really hard time with this question. Suppose that a,b are real numbers with $0<b<a$. Prove by mathematical induction that if $k$ is a positive integer then $a^k-b^k \le ka^{k-1}(a-b)$ 
I know this is a duplicate question that has been answered , but I can't wrap my head around the given solutions.
I especially do not understand how : $a^{k+1} - b^{k+1} = a^k(a-b) + b(a^k-b^k)$
I also tried using the hint : $a^k - b^k = (a-b)(a^{k-1} + a^{k-2}b + ... + b^{k-1}) $ 
I am really lost with this , could someone point me in the right direction please ? 
 A: Hint without Induction: Consider
$$
a^k-b^k=\overbrace{\left(\sum_{j=0}^{k-1}a^jb^{k-1-j}\right)}^{\le ka^{k-1}}\,(a-b)
$$

Hint for Induction:
$$
a^{k+1}-b^{k+1}=a^k(a-b)+\overbrace{\ \ \ \ \ \ b\vphantom{a^k}\ \ \ \ \ \ }^{\le a}\,\overbrace{\left(a^k-b^k\right)}^{\le ka^{k-1}(a-b)}
$$
A: No need for induction to prove this inequality: 
If you've seen the formula for the geometric series, you probably noticed it is deduced from the algebraic (high-school) formula:
$$1-x^n=(1-x)(1+x+\dots +x^{n-1}).$$
Setting $x=\dfrac ba$, you obtain
\begin{align}
a^n-b^n&=a^n\Bigl(1-\frac ba\Bigr)\Bigl(1+\frac ba+\dots\frac{b^{n-1}}{a^{n-1}}\Bigr)\\
&=a\Bigl(1-\frac ba\Bigr)\,a^{n-1}\Bigl(1+\frac ba+\dots\frac{b^{n-1}}{a^{n-1}}\Bigr)\\[0.5ex]
&=(a-b)\bigl(a^{n-1}+a^{n-2}b+\dots+b^{n-1}\bigr)
\end{align}
This formula can be rewritten as
$$a^n-b^n=(a-b)\sum_{\!i+j=n-1}a^ib^j\!\!$$
Now, since $0<b<a$, for each $(i,j)$, $\;a^ib^j\le a^ia^j=a^{n-1}$, and the sum has $n$ terms, so:
$$a^n-b^n\le(a-b)\sum_{\!i+j=n-1}a^{n-1}=(a-b)na^{n-1}$$
A proof by a (direct) induction:
Starting from the relation:
\begin{align}
a^{k+1}-b^{k+1}&=a^k(a-b)+(a^k-b^k)b\\
&<a^k(a-b)+ ka^{k-1}(a-b)b&&\text{by the inductive hypothesis}\\
&<a^k(a-b)+ ka^{k}(a-b)&&\text{since 0<b<a}\\
&=(k+1)a^k(a-b).
\end{align}
A: $a^k-b^k\le ka^{k-1}(a-b).$
Step: $k+1$.
$a^{k+1}-b^{k+1} = $
$a^k(a-b) +b(a^k-b^k) \le$
$a^k(a-b) +bka^{k-1}(a-b)=$
$(a-b)(a^k +bka^{k-1}) \lt $
$(a-b)(a^k +ka^k) =$
$(a-b)(1+k)a^k.$
Used: 
$0 \lt b \lt a$, and 
$a^{k+1}-b^{k+1} = $
$a^k(a-b) +b(a^k-b^k).$
A: Consider function $f(x)=x^k$, $k\in\mathbb{N}$ and segment $[b,a]$, $b,a>0$. Then apply Lagrange's theorem:
$$a^k-b^k=(x^k)'|_{x=\xi}(a-b)=k\xi^{k-1}(a-b),$$
where $\xi\in(b,a)$. Due to monotonic increase of $f'(x)=kx^{k-1}$ one gets
$$a^k-b^k=k\xi^{k-1}(a-b)\leqslant ka^{k-1}(a-b)$$
UPD Sorry, I didn't notice that you have to prove this using induction, I'll leave this answer anyway
A: Actually the part you didn't understand simply means adding and subtracting a term $b a^k$which makes no change to the expression. Then the terms are regrouped.
