With $c=1$ and $n_0=0$ the statement
$(\forall \mathbb{N}_0\ni n>n_0)[n\leq c\cdot (n+1)]$
is obviously true. Does this suffice for the proof of $n=\mathcal{O}(n+1)$?
Another idea was this one:
$\limsup\limits_{n\to\infty}\left|\frac{n}{n+1}\right|=\limsup\limits_{n\to\infty}\left|\frac{\frac{\text{d}n}{\text{d}n}}{\frac{\text{d}(n+1)}{\text{d}n}}\right|=\limsup\limits_{n\to\infty}\left|\frac{1}{1}\right|=1,\;\text{with }0<\limsup\limits_{n\to\infty}\left|\frac{n}{n+1}\right|<\infty\Rightarrow n=\mathcal{O}(n+1)$
Would this suffice?