We all use mathematical induction to prove results, but is there a proof of mathematical induction itself? I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like 
$$1+2+3+\cdots+n = \frac{n(n+1)}{2}$$
However.
I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.)

In my real variable course many years ago we defined the natural numbers (positive integers $\{1,2,\dots\}$ or $\{0,1,2,\dots\}$ by the Peano Postulates/Axioms. The reference we used was Graves L.M. (1946), The Theory of functions of real variables, McGraw-Hill and the postulates were given as follows
We assume the existence of a system $(\mathcal{M},s\}$ where $\mathcal{M}$ is a class of elements $m,n,\dots$, and $s(m)$ may be called the successor of $m$, having the following properties:


*

*$s$ is a function on $\mathcal{M}$ to $\mathcal{M}$; that is to each element in $\mathcal{M}$ corresponds an element $s(m)$ in $\mathcal{M}$.

*$\exists \ m_0 \in \mathcal{M}\text{ and } m_0 \notin s(\mathcal{M})$; that is, there is an element $m_0$ in $\mathcal{M}$ which is not the successor of an element of $\mathcal{M}$.

*$s(m)=s(m) \Leftrightarrow m=n$

*$\mathcal{M}_0 \subset \mathcal{M} \land  s(\mathcal{M}_0) \subset \mathcal{M}_0 \land m_0 \in \mathcal{M_0} \Leftrightarrow\mathcal{M}_0= \mathcal{M}$


Our entire real variable course was built on this set of axioms. One will recognise the fourth axiom as the axiom of induction which is here included in the definition of the integers and thus is assumed and not proven. 
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