There is actually something to prove here. $a+1$ is defined in the Peano axioms as $a+S(0)=S(a+0)=S(a)$. Hence, if $a\in\mathbb{N}$, $a+1=S(a)\in \mathbb{N}$.
We can go further; $a+2=a+S(1)=S(a+1)=S(S(a))$. Hence, if $a\in \mathbb{N}$, $a+2=S(S(a))\in \mathbb{N}$. In fact, for any specific $k$, we can equally prove that if $a\in \mathbb{N}$ then $a+k\in \mathbb{N}$.
There is actually something to prove here. $a+1$ is defined in the Peano axioms as $a+S(0)=S(a+0)=S(a)$. Hence, if $a\in\mathbb{N}$, $a+1=S(a)\in \mathbb{N}$.
We can go further; $a+2=a+S(1)=S(a+1)=S(S(a))$. Hence, if $a\in \mathbb{N}$, $a+2=S(S(a))\in \mathbb{N}$. In fact, for any specific $k$, we can equally prove that if $a\in \mathbb{N}$ then $a+k\in \mathbb{N}$.