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!

$2^{k+1}=2\cdot2^{k}=2^{k}+2^{k}$, 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.
