TITLE: Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$?
QUESTION [5 upvotes]: I need to prove  $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?
This is how I prove this:
Prove:$2n ≤ 2^n$, for all integer $n≥1$
Proof: $2+4+6+...+2n=2^n$
$i.)$ Let $P(n)=1
        P(1): 2(1)=2^1\implies 2=2$.
       Hence, $P(1)$ is true.
$ii.)$ 
Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$
from the assumption add $2(k+1)$ on both sides so we have 
 $2+4+6...2k+2(k+1)=2^k+2(k+1)$
I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$. 
I feel i'm doing something wrong. 
Any help would be appreciated!

REPLY [2 votes]: Recall that, by induction,
$$
2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n-1} + \binom{n}{n}.
$$
All the terms are positive; observe that
$$
\binom{n}{1} = n, \quad \binom{n}{n-1} = n.
$$
Therefore,
$$
2^n \geq n+n=2n.
$$
Remark: I suggest this proof since the plain inductive proof of your statement has been given in many answers. I also believe that an inductive proof of the binomial expansion is an instructive exercise, probably more instructive than the one you are trying to solve.