Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
    0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1 + 1-n = n-n = 0, \forall n \in \mathbb{N}$. Thus we have:
\begin{align*}
    0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a-n)(a-1)\\
    0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a)(a-1)-n(a-1)\\
    0 &\leq a(a^{n-1} + a^{n-2} + \ldots + a + 1)(a-1) -na + n\\
    0 &\leq a(a^{n}-1) -na + n\\
    0 &\leq a^{n + 1} -a-na + n\\
    0 &\leq a^{n + 1} -(n + 1) a + n\\
    (n + 1)a &\leq a^{n + 1} + n\\
\end{align*}
Another form of reason is using the inequality of the arithmetic mean with the geometric mean, as follows:
\begin{align*}
    \frac{a^{n + 1} + n}{n + 1} &= \frac{a^{n + 1} + 1 + 1 + \ldots + 1 + 1}{n + 1}\\
                          &\geq \sqrt[n + 1]{a^{n + 1}(1)(1)\ldots(1)(1)}\\
                          &= \sqrt[n + 1]{a^{n + 1}}\\
                          & = a.
\end{align*}
Finally it is solved and we have $a^{n + 1} + n \geq (n + 1)a$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

Alternatively, the result follows from the Bernoulli’s inequality:
$$[1+(a-1)]^{n+1}\geq 1+(n+1)(a-1),$$ which is true for all real numbers $a\geq 0$, and all integers $n\geq -1$. (Exclude the case when $a=0,n=-1$ if necessary to avoid the ambiguity $0^0$.)
