Help with Elementary number theory please Use the second principle of finite induction to establish that for all $n\geq1$ :
$$a^n-1=(a-1)\left(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1\right) $$
Step by step explanation please! I'm confused how the second principle of finite induction is different from the first. 
 A: You can use that
\begin{align}
a^{k+1}-1&=a(a^k-1)+(a-1)\\
&=a(a-1)(a^k+\cdots+a+1)+(a-1)\\
&=(a-1)(a^{k+1}+\cdots+a^2+a)+(a-1)\\
&=(a-1)(a^{k+1}+\cdots+a+1)
\end{align}
assuming you know that the theorem holds for $n=k$, but this is just simple induction (I think), not the so called 'second principle' of finite induction.
To answer your question:
The second principle uses all cases $n=1$ up to $n=k$ as induction hypothesis, instead of only $n=k$. See this answer on a similar question for more information.
EDIT
A quick google search returned the page above, which uses the similar wording. I had never heard of it (the phrasing 'second principle' instead of 'strong induction') myself too.
A: Second principle of finite induction establishes if the given statement is true for $n$ from $1,2,3... n$. Then it is true for $n+1$.
$$a^1 -1= a-1 \\
a^2 -1= (a-1)(a+1)$$
say $n \ge 3$
assume
$$a^{n} -1 =(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$$
we have to prove this for $n+1$.
$a^{n+1} -1$ can be written as $(a+1)(a^{n-1})-a(a^{n-1} -1)$
we already know $a^n -1$ is $(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$
Let $S=a^{n-1} +a^{n-2} +a^{n-3}.....+a+1$
so I can write $a^{n+1} -1$ as $(a+1)(a-1)S -a(S-a^{n-1})(a-1)$
which on simplifying gives
the required statement to prove
Hope it helps.
