Is my induction proof of $2^{n} > 2n+1$ correct? Hello I am wondering if anyone can conform that the method I use in the following proof is valid. If not please inform me/ point me in the right direction.
It is a very basic question, i.e. to prove that for $$n \ge 3$$ $$2^{n} \gt 2n+1$$
My approach.
The base case is clear because $8 \gt 7$
Now, suppose the statement is true for $n=k$ i.e. that $$2^{k}\gt 2k+1$$
and we must show now that $$2^{k+1} \gt 2(k+1)+1=2k+3$$
My thought was that this can be shown to be true if $$2^{k+1}-(2k+3) \gt 0$$
$$2(2^{k})-(2k+3) \gt 2(2k+1)-(2k+3)=2k-1 \ge 5 \gt 0$$
hence it holds.
Would that be sufficient?
Thanks
 A: What you did, nothing was wrong in it. 
Another approach:
$2^{k+1}=2\cdot 2^k\gt 2(2k+1)=2k+2k+2\ge2k+8\gt2k+3$,as you wished.
A: There are only two things I would comment on, one technical and one stylistic:


*

*"Suppose the statement is true for $n=k$." What is $k$? Are you assuming the statement is true for any $k$ (i.e., does $k$ vary or is it fixed?)? Do we have any constraints on $k$? Etc. 

*Stylistically, most induction proofs are written by moving from the left-hand side of $S(k+1)$ (where $S$ is the statement in question) to the right-hand side of $S(k+1)$, and there is good reason for this proof structure and should only be abandoned, I think, when really necessary. Such a structure preserves clarity among other things, and there is really not a reason to do away with that structure in this problem. You may find it helpful to read this post on how to write a clear induction proof. Below, I'll write out a proof for your problem that uses the style provided in that link.

Claim: For all $n\geq 3, 2^n>2n+1$. 
Proof. For any integer $n\geq 3$, let $S(n)$ denote the statement
$$
S(n) : 2^n>2n+1.
$$
Base step ($n=3$): $S(3)$ says that $2^3=8>7=2(3)+1$, and this is true.
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 3$. Assume that
$$
S(k) : 2^k>2k+1
$$
holds. To be proved is that
$$
S(k+1) : 2^{k+1}>2(k+1)+1
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
2^{k+1} &= 2\cdot 2^k\tag{law of exponents}\\[0.5em]
&> 2\cdot(2k+1)\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&= 2k+(2k+2)\tag{manipulate}\\[0.5em]
&> 2k+(2+2)\tag{since $k\geq 3$}\\[0.5em]
&> (2k+2)+1\tag{since $2>1$}\\[0.5em]
&= 2(k+1)+1,\tag{factor out $2$}
\end{align}
one arrives at the right-hand side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.
By mathematical induction, it is proved that for all $n\geq3$, the statement $S(n)$ is true. $\blacksquare$

Wouldn't you agree that argument was much clearer than your original argument? It's not that yours was wrong (save the lack of explanation whether or not $k$ varied for your inductive assumption). There's simply more of an art form to framing/writing induction proofs than most proofs. 
