Clarification on inductive proof of Bernoulli's inequality 
Prove that if $h > -1$, then $1 + nh ≤ (1+h^n)$ for all nonnegative integers $n$.

I've read several solutions and I'm still totally lost on how to go about this. 
I have the inductive hypothesis: $1+kh ≤ (1+h)^{k}$.
And that we want to prove: $1+(k+1)h ≤ (1+h)^{k+1}$.

From here, I found various first steps that I don't understand. Examples:


*

*$(1+h)(1+kh) ≤ (1+h)^k(1+h)$. How did they get the left side?

*$1 + kh + h ≤ (1 + h)(k + kh)$. How did they arrive at the right side?
 A: Note: 
An inductive proof is given as an answer by David W.Farlow.

Explanations of some steps and flaws in your inductive argument:


*

*We get $(1)$ from the inductive hypothesis (which is assumed to be true.) $$(1+kh) \leq (1+h)^k \Rightarrow (1+h)(1+kh) \leq (1+h)^k(1+h)=(1+h)^{k+1}$$ as $h >-1 \Rightarrow h+1>0$ and so multiplication of both sides of an   inequality with a non-negative number doesn't change the inequality.

*Here, $(2)$ is false because $$1+k+kh \leq (1+h)(k+kh)=k+2hk+h^2k \Rightarrow 1 \leq hk+h^2k=hk(h+1)$$ In this case, $h>-1$ but the other factors are non-negative; so it can't be true when $-1<h<0$.

*So, $(2)$ should be $$1+k+kh \leq (1+h)(1+kh)=1+h+kh+kh^2$$ which is true because $h^2k \geq 0$


Now, why do we do $(1)$ and $(2)$ and how do we get the LHS or RHS in various inequalities?

This is done because to prove $a \leq b$, if we get $a \leq c$ and $c \leq b$, then we get the required inequality. Accordingly we can choose the required value of $c$ which not only makes this task possible but also easy.
A: First of all, why have you been reading a bunch of solutions? I suggest you make a serious attempt at the problem yourself and only then look to other solutions to bridge whatever gap that exists in your reasoning (looking up a solution should be the very last resort). To that end, note that the inequality is trivial for when $n=0,1$. Thus, consider the following statement of your problem (which I will provide a proof for afterwards):

Problem: For any non-zero real number $x$ with $x>-1$, prove by induction that for each $n\geq2$, $$(1+x)^n>1+nx.$$

Proof. Fix $x\in\mathbb{R}_{>-1}$, where $x\neq0$, and let $S(n)$ denote the statement $S(n) : (1+x)^n>1+nx$. Because $x\neq0$, we know that $x^2>0$, and hence $(1+x)^2=1+2x+x^2>1+2x$, whereby we see that $S(2)$, the base case, holds. Now, fix $k\geq2$, and assume that $S(k)$ holds. Then
\begin{align}
(1+x)^{k+1}&=(1+x)(1+x)^k\tag{$x>-1$; exponent law}\\[1em]
&> (1+x)(1+kx)\tag{by $S(k)$, then ind. hyp.}\\[1em]
&= 1+kx+x+kx^2\tag{expand}\\[1em]
&> 1+kx+x\tag{since $kx^2>0$}\\[1em]
&= 1+(k+1)x.\tag{factor}
\end{align}
Thus, $S(k+1)$ is also true, completing the inductive step. $\blacksquare$

Does it all make sense now? The margin notes for the proof should help a great deal (indeed, they should answer both of the questions you had about different points in other proofs). Regardless, always try as many approaches as possible and only then (after numerous fruitless attempts) give recourse to an externally produced solution. 
