Help with proof of contrapositive of well-ordering principle 
Prove by induction on $n$ that if $A$ is a set of positive integers without a least element, then $\mathbb{N}_n \subseteq \mathbb{Z}^+ - A$ for every $n$ so that $A$ is the empty set.

I don't understand the inductive step. The solution provided is:

Suppose as inductive hypothesis that $\mathbb{N}_k \subseteq \mathbb{Z}^+ - A $ for some positive integer $k$. Then if $k+1 \in A$ it would be the least element by inductive hypothesis, which is impossible. Hence $k+1 \in \mathbb{Z}^+ - A$ and so, using the inductive hypothesis again, $\mathbb{N}_{k+1}\subseteq \mathbb{Z^+}-A$, completing the inductive step.

I don't understand the second sentence onwards. I do understand that if $k+1 \in A$ then $k+1$ will be the least element of $A$ (definition of least element) but I don't see how this follows from the induction hypothesis. I also don't get how to use the induction hypothesis a second time again to reach the conclusion  $\mathbb{N}_{k+1}\subseteq \mathbb{Z^+}-A$. Furthermore, why do we also not consider the case $k+1 \notin A$?
Here's how I currently see thing:
$P(k): \text{A is a set of positive integers without a least element } \Rightarrow  \mathbb{N}_{k}\subseteq \mathbb{Z^+}-A$
$P(k+1):\text{A is a set of positive integers without a least element } \Rightarrow  \mathbb{N}_{k+1}\subseteq \mathbb{Z^+}-A $
I need to prove $P(k)\Rightarrow P(k+1)$
So, if I assume $P(k)$ is true, how does it follow that $k+1$ is the least element in $A$? All I can see is that $A$ cannot have a least element because $P(k)$ says so. 
 A: I would say that the problem and the solution are not framed in the clearest possible way.  
Problem (Contrapositive of WOP): If $A\subseteq \mathbb{Z}^+$ and $A$ has no least element, then $A=\emptyset$.
Proof (by 'strong' induction): We'll show that for every positive integer $n$, $n\notin A$.
Basis step: $1\notin A$ for otherwise $1$, being the least element of $\mathbb{Z}^+$, would automatically be the least element of $A$, but this contradicts $A$ having no least element.
Induction hypothesis: Assume for some positive integer $k$ assume that $1, 2, ..., k\notin A$.  That is $A\subseteq \{k+1,k+2,\dots\}$
Inductive step: $k+1 \notin A$ for otherwise by the induction hypothesis, $k+1$ would be the least element of $A$, but this contradicts $A$ having no least element..
A: Assume $P(k)$ is true, and assume $A \subset \mathbb{Z}_+$ has no least element.
Then $k+1$ is not the least element of $A$. But since $0, \dots, k \notin A$ this means that $$ \forall n \in A,\  \  n \notin \{ 0, \dots, k\}$$
i.e.
$$ \forall n \in A,\  \  n \geq k+1$$
so $k+1 \notin A$ (otherwise $k+1$ would be the least element of $A$).
Hence $0, \dots, k, k+1 \notin A$, and we proved $P(k+1)$.
A: your statement $P(k)$ is equivalent to $(0 \notin A)\land (1 \notin A) \land \cdots \land (k \notin A) $
if you assume $P(k)$ then $\lnot P(k+1) \implies \lnot((k+1) \notin A)$ i.e. $(k+1) \in A$. call this element $a$. 
if there were $a^* \in A, a^* \lt a$, then $a^* \le k$ implying $\lnot P(k)$ contrary to hypothesis. 
so $a$ is the least element of $A$. but this is also contrary to hypothesis. so we must have $P(k+1)$
