The following is my understanding of how mathematics works, please correct anything below that is false/misguided:
The induction theorem, used to prove statements that apply over the set of natural numbers, is generally constructed from the axioms of set theory (eg. ZFC). From there you can construct other sets that derive from the natural numbers ($\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$, etc.) I only have a rudimentary knowledge of ZFC and the construction of such sets.
However, I believe that the properties and operations of the real numbers can also be proven using only logic and these axioms in a 'synthetic' approach. Recently I have been trying, using this synthetic approach, to prove various theorems of the real numbers. When it comes to exponentiation, which has a recursive definition, it seems that I have to use induction to prove various things (for example, to prove that exponentiation is a function). But to use induction, I need to prove the induction theorem from these 'synthetic' axioms. Can this be done, or is it only possible to prove such things by constructing everything using set theory as the foundation? If it can be done, how?
What you have to realize is that you still need some set theory (I think) to do this. One reason is that the concept of fields are defined in terms of sets (a field $(\mathbb R, +, \cdot)$ is a set $\mathbb R$ together with...). You will also probably need second order logic to state the existence of $\mathbb N$.
Also you have to realize that no matter how you do it it's a lot of work ahead, basically the same amount as if you start at $\mathbb N$. The difference is that you go the other way around - instead of extending $\mathbb N$ you identify $\mathbb N$, $\mathbb Z$, $\mathbb Q$ as subsets of $\mathbb R$. For this reason I will not go into details, but rather scetch the approach.
The big question is how to identify $\mathbb N$ since that set has a central role in the induction principle. The answer lies in the that it basically has do fulfil the axioms of Peanno, but we don't have to be that primitive now.
What we do is to consider (associative) submonoids under addition that also contain the multiplicative identity (they contain the additive identity too, by definition). We observe that such submonoids exists as $\mathbb R$ is one. We construct $\mathbb N$ as being the intersection of all these (underlying sets).
Now there's a number of steps that is required to prove the induction principle. We note that since $\mathbb N\subseteq\mathbb R$ we can use the operations from $\mathbb R$ on $\mathbb N$ (although we haven't proven that $\mathbb N$ is closed under these (in fact it isn't closed under subtraction and division).
If there weren't we could form a monoid by excluding $x$
If we have a monoid we still get a monoid if we remove all elements between $0$ and $1$ and the negative elements.
This is showed by dividing the set greater and less integers. We see that at most two integers we have $|x-k|<1$. The infimum must be the smaller of these and can't be zero because $x\notin N$.
This is shown by using the $\sup$ ($\inf$) and use proposition 3 to show that it's inside the subset.
We use proposition 2 to show that in proposition 1 $a$ and $b$ is between $0$ and $x$. We then use proposition 4 to find a minimal $b$ that fulfills the equation $x=a+b$. We use proposition 1 to show that this $b$ must be $1$.
Now we're ready to show the induction principle:
Consider $L = \mathbb N\setminus K$. If $L\ne\emptyset$ it will be bounded below by $0$ because it's a subset of $\mathbb N$. So there exists a minimal element $n\in L$. Now if $n=0$ it would contradict that $0\in K$. Otherwise we have from proposition 5 that $n=a+1$ for some $a$ and since $a<n$ we have $a\notin L$ so $a\in K$ and by assumtion $n=a+1\in K$ which is also a contradiction. Therefore we conclude that $L=\emptyset$ which means that $K = \mathbb N$.
As for the rest of the axioms of Peanno we see that the axioms 1-5 follows directly as they are true for any field. For the axioms 6-8 we define $S(n) = n+1$ and they're quite straight forward to prove.
After this one can take about the standard approach in building up from Peanno's axioms, but things are a bit easier since we don't have to do "funny constructs". We can for example define rational numbers as the quotient of integers (since we already have a field).
As for recursively defined functions (on $\mathbb N$) that is an allowed construct. When you prove that however you should not use notation that suggests that it is in fact a function, not before the proof is complete. For example if we take exponention we will define it as $x^0 = 1$ and $x^{n+1}=x^nx$, but before the proof we will use the relation $E$ defined as $(x, 0)Ey \leftrightarrow x=y$ and $(x,n+1) E z \leftrightarrow \exists y: (x,n) E y\land z=xy$. We then by induction (over $n$) prove that for every $x\in\mathbb R$ and every $n\in\mathbb N$ we have exactly one $a$ such that $(x,n)Ea$ which makes $E$ a function. The proof is somewhat straight forward, but some attention has to be paid for the case where $x=0$ (in which case $z=0y=0$ regardless of $y$ which therefore exists). After this we introduce the expression $x^n$ as $(x,n) E x^n$.
Then one goes on to prove the standard rules for exponentiation by induction. Then these rules makes it natural to define $x^{-n}$ as $1/x^n$ and it's defined for integers. Now we reprove the standard rules for exponentiation (but this time we don't need induction). The next steps is a bit tricky, but we define $x^{p/q} = a$ if there's a number $a$ such that $x^p = a^q$ - we use completeness to prove that such a number always exists (except for the exceptions). The final part is to define $x^a$ for arbitrary $a$ which is again done via completeness. We have to reprove the rules of exponentiation for each of these two steps too.