Prove that $(1+x)^n>1+nx$ for all integers $n\geq2$, where $x>-1$ and $x\neq0$. I would like to get feedback on whether this proof is valid or not and would like to know if my usage of set-builder notation is correct.
I get the feeling that the discrete mathematics class I'm in is relatively dumbed down as my professor doesn't really go over this stuff. However, I'm interesting in pursuing more advanced mathematics in future studies and don't want to wait to learn this stuff.
Thanks in advance!
Proof (by mathematical induction):
Suppose $\{x\in\mathbb{R}\mid x\gt-1\land x\neq0\}$, and let the property $P(n)$ be the inequality
$$(1+x)^n\gt1+nx\text{.}$$

^^^ For example, is my use of set-builder notation correct up here? ^^^

It will be shown that $\forall\{n\in\mathbb{Z}\mid n\geq2\}$, $P(n)$ is true. This will be done by mathematical induction on $n$.
Show that $P(2)$ is true:
$P(2)$ is the inequality
$$(1+x)^2\gt1+2x\Rightarrow1+2x+x^2\gt1+2x\Rightarrow x^2\gt0\text{.}$$
$\therefore P(2)$ is true $\because x\neq0$.
Show that $\forall\{k\in\mathbb{Z}\mid k\geq2\}$, $P(k)\rightarrow P(k+1)$:
Let $\{k\in\mathbb{Z}\mid k\geq2\}$, and suppose that
$$(1+x)^k\gt1+kx\text{.}$$
$1+x\gt0\because x\gt-1$. Hence
$$(1+x)^k(1+x)\gt(1+kx)(1+x)\Rightarrow(1+x)^{k+1}\gt1+(k+1)x+kx^2\text{.}$$
$kx^2\gt0\because k\geq2\land x\neq0$. $\therefore$ it follows that
$$(1+x)^{k+1}\gt1+(k+1)x\text{.}$$
Hence $P(k+1)$ is true. QED.
 A: To begin with: The logic of your proof is all correct. As you asked for it, I will give some hints rather about the notatin and overall writing style of a mathematical proof.
Yes, $\{\,x\in\Bbb R\mid x>-1\land x\ne 0\,\}$ is correct set-builder notation. However, it denotes a set whereas syntactically in "Suppose ...", you need a statement. Hence you might write 

Let $S=\{\,x\in\Bbb R\mid x>-1\land x\ne 0\,\}$. Suppose $x\in S$, and let the property ...

However, unless you need to repeatedly reference the set $S$ further down in your proof, I'd either use "pre-fabricated" sets, such as

Suppose $x\in \left]-1,\infty\right[\setminus\{0\}$ and let ...

or why not simply and most readably

Suppose $x>-1$, but $x\ne 0$. Let $P(n)$ ...

Likewise, you cannot say $\forall \{\,n\in\Bbb Z\mid n\ge 2\,\}$. For a formal statement, you might write
$$ \forall n\in\{\,n\in \Bbb Z\mid n\ge 2\,\}\colon P(n)$$
or less convoluted
$$ \forall n\colon n\in\Bbb Z\land n\ge 2\to P(n)$$
plus a few similar variants. Actually, since we are going to use induction, I'd prefer to write use $\Bbb N$ in place of $\Bbb Z$ here to make the notation more suggestive, perhaps
$$ \forall n\in \Bbb N\colon n\ge 2\to P(n).$$ 
But ultimately, you use this not as a formal statement but as parts of speech in a natural language sentence. Pleas avoid using math formula fragments where conjunctions etc. are needed. If you do this, $\implies$ your text ia less pleasent $\forall$ readers $\land$ they will $\neg$ like it. Why not

It will be shown that for all integers $n\ge 2$, $P(n)$ is true.

Your proof of the base case uses the implication symbol in the wrong direction. What you want to show is that $(1+x)^2>1+2x$, but what you write is that this desired inequality implies ("$\Rightarrow$") the obviously troe fact $x^2>0$. While the order you write things in are the way you find the proof, it is not the way you can present the proof. If you can justify that "$\Rightarrow$" is in fact "$\Leftrightarrow$", then you are fine. Otherwise or perhaps anyway, it is better tohave the desired result at the end, a la

As $x$ is real and $\ne0$, we have $x^2>0$, hence
  $$ (1+x)^2=1+2x+x^2>1+2x,$$
  as desired.

All of the above remarks apply also to various places in your proof of the induction step.
