Proving $n \lt 2^n$ for $n\geq 1$ using induction Very close to understanding this, hopefully. Via induction, I'm following a proof but can't understand one of the last steps.
Claim: $n < 2^n$ for natural numbers $n = 1, 2, 3,\ldots$ 
For step one, $n = 1$, this is obviously true:
$$1 < 2^1 = 1 < 2.$$      
Next, assuming $n = k$, is true:
$$k < 2^k.$$            
Next I must show that 
$k + 1 < 2^{k + 1}$ is true, to prove that all natural numbers are true. I begin with this:
$$
k < 2^k\\
k + 1 < 2^k + 1
$$
Since this is true, adding one more to the RHS is also going to be true:
$$k + 1 < 2^k + 2.$$
Here's where I don't get it. The proof I'm reading claims it's obvious that
$$2^k + 2 \leq 2^{k+1}.$$
Ok yeah, it seems like this is true, but are we certain? Substituting $k = 1$ and $k = 2$ does the trick, and it does seems reasonably intuitive that this would go on forever, but how is this a "formal" result?
From there the proof is completed by putting everything side by side, showing that $k + 1$ is less than all those intermediate steps, resulting in it also being less than $2^{k + 1}$, which makes sense, had I understood that last step! 
Yes there are a number of questions similar to this one - couldn't quite find one for this particular glitch. Much thanks!
 A: $2^{k+1}=2\cdot2^{k}=2^{k}+2^{k}$, which will be greater than $2^{k}+1$ so long as $2^{k}>1$. This is true for all values of $k$ you're considering.
A: Oddly enough, one of the most difficult things about induction problems is actually writing a clear induction proof. As such, I will provide a proof in the spirit of the template I provided a link to above. Hopefully it will clear up any confusion you may have on the matter.

Claim: For all $n\geq 1, n<2^n$.
Proof. For any integer $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : n<2^n.
$$ 
Base step ($n=1$): $S(1)$ says that $1<2^1$, and this is true. 
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$. Assume that
$$
S(k) : \color{green}{k<2^k}
$$ 
holds. To be shown is that
$$
S(k+1) : \color{blue}{k+1<2^{k+1}}
$$
follows. Beginning with the left side of $S(k+1)$, 
\begin{align}
\color{green}{k}+1 &< \color{green}{2^k}+1\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&< 2^k+2^k\tag{since $1<2^k$ for all $k\geq 1$}\\[0.5em]
&= 2\cdot 2^k\tag{group like terms}\\[0.5em]
&= \color{blue}{2^{k+1}}\tag{exponent law}
\end{align}
one arrives at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step. 
Conclusion: By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
A: Another approach: $2^1=2 >1 $ and left hand side grows faster than R.H side after $x=1/2$, i.e., $ \frac {d}{dx} 2^x = \frac {d}{dx} (e^{xln2})=ln2(2^x)> \frac {d}{dx}(x)=1$ for $n>1/2$
So $2^1>1 $ , and $2^n$ grows faster than $n$ after $n=1$ ( after around $x=1/2$).
