Prove by induction that $n!+n \geq 2^n$

Question

I'm trying to use the principle of mathematical induction to prove that $n!+n\ge2^n$ for all $n\ge1$. I know how to prove $n!\ge2^n$, but I can't figure out what to do with the extra $n$. For instance, if I try to factor $n+1$ out of the inductive case,

$(n+1)!+(n+1)\ge2^{n+1}$

$(n+1)(n!+1)\ge2\times2^n$

I'm left with $n!+1$, which I don't know what to do with. Similarly, if I try to use multiplicative telescopy, which I used to prove $n!\ge2^n$, I get stopped by the random $n$ that I can't multiply.

Any help would be greatly appreciated.

Answer 1

Clearly $n!+n>n!$, so if you already could show that $n!\ge 2^n$ (at least for $n\ge2$), then you need only check $n!+n\ge 2^n$ for the remaining case $n=1$.

P.S.: Sometimes showing a stronger result is easier.

Answer 2

If you know from induction that $n! \geq 2^{n}$, and $n!+n \geq n!$ for every $n$, then you have the inequality you want, right?

Answer 3

Case $n = 1$: $$ 1! + 1 = 2 \ge 2^1 $$ Assuming $n$ holds, checking the statement for $n+1$: $$ \begin{align} (n+1)! + (n + 1) &= (n+1) n! + n + 1 \\ &= 2 \underbrace{(n! + n)}_{\ge 2^n} + (n-1) n! - n + 1 \\ &\ge 2^{n+1} + (n-1) n! - n + 1 \\ &\ge 2^{n+1} + \underbrace{(n-1)}_{\ge 0} \underbrace{(n! - 1)}_{\ge 0} \\ &\ge 2^{n+1} \end{align} $$ The statement is then true for $n+1$ as well and by principle of induction true for all $n \in \mathbb{N}$.