Prove by mathematical induction $4^n > n+1$ Prove the following by mathematical induction: 
$4^n > n+1$, for all integers $n ≥ 1$
Step 1:
$n=1$:
LHS $= 4^{(1)} = 4 $
RHS $= (1) + 1 = 2$
LHS > RHS. ∴ $P(1)$ is true.
Step 2:
Assume $P(k)$ is true for some $k ≥ 1$
$P(k)$: $4^k > k+1$
Step 3: We must show $P(k+1)$ is true.
$n = k+1$: $4^{k+1} > (k+1)+1$
RHS = $(k+1)+1 < 4^k +1 <  4^k + 4^k = 2*4^k < 4*4^k = 4^{k+1}$ = LHS
Hence, $P(k+1)$ is true. Therefore, By Math. Induction $P(n)$ is true for all $n ≥ 1$
Can anyone check if my method is correct or there is a better way to do it. Thank you.
 A: This looks all right. I think you could have been a little bit more neat when writing it down, perhaps laying your reasoning vertically as it is usually done, and with a little bit more info. Like this.
Our inductive hypothesis is $4^k>k+1$. We want to prove $4^{k+1}>(k+1)+1$.
Because of our inductive hypothesis,
$4^{k}+1>(k+1)+1$
From this follows
$4*4^k>4^k+1>(k+1)+1$
$4^{k+1}>(k+1)+1$
Yes it takes more time but it makes everything easier for anyone correcting your proof. Anyhow, congratulations! Your proof is correct.
A: Agreed, your proof's fine. If you want a "better" way to do it, I can only suggest something that's "better" in the sense of making the calculation shorter. Note the inductive hypothesis implies $4^{k+1}>4k+4>k+2$.
When dealing with integers, I often prefer to write $a>b$ as $a\ge b+1$. In this case, the theorem itself would be restated as $4^n\ge n+2$, with inductive step $4^k\ge k+2\implies 4^{k+1}\ge 4k+8\ge k+3$. But, as you can see, I'm really fishing for "improvements" at this point because it's basically fine as is.
