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The following is from Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra; Edited by H. Behnke, F. Bachmann, K. Fladt, W. Suess and H. Kunle

I've used square brackets '[' and ']' to indicate where referenced text is copied in place.

The quoted text is part of the development of the theory of natural numbers. Their version of Peano's axioms begins with 1, not 0. I am able to follow the discussion up to the proof of rule (15).

If for the numbers $a,b$ there exists a number $c$ with $a+c=b,$ we write $a<b$ ($a$ is less than $b$), or alternatively $b>a$ ($b$ is greater than $a$). For the relation $<$ defined in this way we have the following theorems:

12. if $a<b,$ then $ane b$ (antireflexivity);




13. if $a<b$ and $b<c,$ then $a<c$ (transitivity);




14. if $a<b,$ then $left(a+dright)<left(b+dright)$(monotonicity of addition);




15. if $ane b,$ then $a<b$ or $b<a.$

Rule (12), which states that $a+cne a$ for all $a,c,$ is proved by complete induction on $a,$ for we have $1+cne1$ by (1) [$a+1=a^prime$],
(7) [$a+b=b+a$] and axiom IV [$a^primene1$ for every number $a$]; and if we had $a^prime+c=a^prime,$ it would follow that $left(a+cright)^prime=a^prime,$ and thus $a+c=a.$ For
the proof of (13), (14) we set $a+u=b,b+v=c$ and thus get

$c=left(a+uright)+v=a+left(u+vright),$

$b+d=left(a+uright)+d=a+left(u+dright)=left(a+dright)+u.$

Complete induction on $a$ is again used to prove (15), as follows.

The case $a=1$ is first dealt with by complete induction on $b:$
[footnote: In this case the induction hypothesis is not used at all, a fact which may make the proof somewhat harder to follow.]

$1=1;$

$1<1+b=b+1=b^prime.$

Then from (15) (for all $b$) the same statement with $a^prime$instead of $a$ (thus for all $b:$ $a^prime<b$ or $a^prime=b$ or
$b<a^prime$) is derived by complete induction on $b:$

$1<a^prime;$

$a^prime<b^prime$ or $a^prime=b^prime$or $b^prime<a^prime$
by (15) and (14);

here again the induction hypothesis ($a^prime<b$ or $a^prime=b$ or $b<a^prime$) is not used.

Since they define complete induction as proving a proposition $Pleft[nright]$ true for all $ninmathbbN$ if an only if $Pleft[1right]$ and $Pleft[nright]impliesPleft[n^primeright],$ I'm not sure what it means to not use the induction hypothesis $Pleft[nright]$ in a proof by complete induction?

Will someone please explain this to me?

Yes, you have to prove $P[n] Rightarrow P[n']$, where $P[n]$ is the hypothesis, but sometimes you can show $P[n']$ without using the hypothesis $P[n]$. But of course, in showing $P[n']$, you have automatically shown $P[n] Rightarrow P[n']$, so you still show what is need for the inductive proof.

To be concrete: in their proof they are tring to show that is $a neq b$, then $a<b$ or $b<a$ for any $a$ and $b$. To prove this, they first consider $a=1$, so now they need to show that for all $b$: if $b neq 1$, then $1<b$ or $b<1$. And for that, they use induction:

Base case: $b=1$. Well, then $b neq 1$ is false, making the whole conditional if $b neq 1$, then $1<b$ or $b<1$ true. Done with base.

Step: Suppose we have if $b neq 1$, then $1<b$ or $b<1$ for some specific $b$ (this is the inductive hypothesis). Now we need to show that if $b' neq 1$, then $1<b'$ or $b'<1$. Well, we know that $b'1=b+1=1+b>1$, so done ... but note: in showing that if $b' neq 1$, then $1<b'$ or $b'<1$, we never used the hypothesis that if $b neq 1$, then $1<b$ or $b<1$. In fact, we also didn't use $b' neq 1$.

We directly showed:

$$1<b'$$

,therefore

$$1<b' text or b'<1$$

,therefore

$$textif b' neq 1 text then 1<b' text or b'<1$$

, and therefore

$$[textif b neq 1, text then 1<b text or b<1] text then [textif b' neq 1 text then 1<b' text or b'<1]$$