Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$.

Define $f(x)=x^{n+1}-(n+1)x+n$. Note that $f(1)=0$ and $f'(x)=(n+1)(x^n-1)$. Therefore, $f'(x)$ takes negative values in $(0,1)$ and positive values in $(1,\infty)$, as desired. So $f(x)$ takes its minimum value (which is zero) at $x=1$.

I found another way to solve it, by Taylor series for $a^x$, we have: \begin{align*} a^x\geq 1+x. \end{align*} As long as $a>1$. Now if $x = y-1$, we have: \begin{align*} a^{y-1}&\geq 1+(y-1)=y\\ a^{y-1}&\geq y\\ a^{y}&\geq ya \end{align*} Thus, if $y = n+1$, \begin{align*} n+a^{n+1}\geq a^{n+1}\geq (n+1)a. \end{align*} Finally for $ a = 1 $ we have trivially.