Question regarding Strong Principle of Induction I'm currently studying Discrete mathematics from a book by Normal L. Biggs and i don't understand the thinking about an example on Strong Principle of Induction,
The example i need help understanding:

Prove that 
  $ 
n^2 > 7n +1, \forall  n \ge 8
$
Solution
The result is true when n=8 because $8^2 =64$ and  $ 7 \times  8 = 57 $. 
  Suppose it is true when $ n $ is any number $ k \ge 8 $, that is $ k^2 > 7k + 1 $.
$ (k + 1)^2 = k^2 + 2k + 1 > (7k + 1) + 2k + 1 = 7(k + 1) + 1 + (2k-6) $
Since $ k \ge 8, 2k - 6$ is an natural number and the last expression above is greater then $ 7(k+1) + 1 $. The induction is verified, and so the result is true for all $ n \ge 8 $

And here comes the part i don't understand.
I don't understand, shouldn't it be $ (k+1)^2 > 7(k+1) + 1 $?
Where does the the "$2k$" come from in $ (7k + 1) + 2k + 1 $ and how does it result in the prof? Could someone explain on a basic level on the thinking behind this example?
 A: Because we have :

$(k+1)^2=k^2+2k+1$

and, by induction hypotheses we have :

$k^2 > 7k+1$.

Thus, "substituting" the inequality in the first equality, we get :


$(k+1)^2 > [7k+1]+2k+1$.


A: Study Induction again !
In this step you just know for every $l\leq k$ : $l^2>7l+1$.
Here one doesn't know anything about $k+1$ !
Then we have $(k+1)^2=k^2+2k+1$ and apply "induction hypothesis" on $k$.
A: Alright, so we already have the base case ($k=8$), so we need to prove the inductive hypothesis.
We assume that 
$$k^2 > 7k +1$$
for some $k\ge8$: we now want to show that 
$$(k+1)^2 > 7(k+1) + 1.$$
We know that $(k+1)^2 = k^2 + 2k + 1 > (7k+1) + 2k + 1$.  We can simplify this last expression to $9k + 2 = 7(k+1) + 1 +(2k - 6)$.  As the example remarks, $k\ge8$, so $(2k-6) > 0.$  Thus,
$$(k+1)^2 > 7(k+1)+1 +(2k-6) > 7(k+1) + 1 + 0,$$
or more simply,
$$(k+1)^2 > 7(k+1) + 1,$$
as was to be shown.
I'll also remark that this example only requires weak induction, since we only need to assume the $k$th holds to prove that the $(k+1)$th case also holds.
