Let the subset of elements $n$ in $\boldsymbol{N}$ for which $n \neq n^{+}$ holds be denoted as $S$. By axiom (ii), we know $e \neq e^{+}$, so $e \in S$, and $S$ is non-empty. If $n \in S$, i.e., $n \neq n^{+}$, we will prove that it must be the case that $n^{+} \in S$. If not, then we would have $n^{+}=\left(n^{+}\right)^{+}$. From this and axiom (iii), it follows that $n=n^{+}$, which is a contradiction. Therefore, by the induction axiom (iv), we conclude $S=N$. Proof completed.