let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$ I tried to prove this by induction:
Base case for n = 1 satisfies $\because$ $1+x$ = $1+x$
I.H: $(1+x)^k \geq 1 + kx$
Inductive Step for $k+1$:
$(1+x)^{k+1} \geq 1 + (k+1)x$
$(1+x)^k (1+x) \geq 1 + kx + x$
$(1+kx) (1+x) \geq 1 + kx + x$ by I.H.
$kx^2 + kx + x + 1 \geq 1 + kx + x$
$kx^2 \geq 0$ which is true for $x>-1$ therefore proved?
Is this the correct way to prove by induction?
 A: The idea here is correct, but the way it is written, it's not clear that this is a proper proof of the inductive step. When proving by induction, you start with the $n=k$ case, and then prove the $n=k+1$ case. If we are to assume the implications are top implies bottom, then what you have done is started with the $n=k+1$ case and then worked down. However, this is precisely what you are trying to prove to begin with.
An example of when this kind of logic doesn't work:
Suppose we want to prove $1=0$. We have
$$\underset{\color{red}X}{(1=0)}\Rightarrow\underset{\color{green}\checkmark}{(0\cdot1=0\cdot0)}\Rightarrow\underset{\color{green}\checkmark}{(0=0)}$$
While the conclusion is in fact true, it does not imply $1=0$.

Instead, start with
$$(1+x)^k\ge1+kx$$
and then work towards proving $(1+x)^{k+1}\ge1+(k+1)x$ from there.
A: You can save your proof  by added "is implied by" ($\Leftarrow$) marks between lines to indicate the desired direction:
Inductive Step for k+1
:
$(1+x)^{k+1}≥1+(k+1)x\Leftarrow$
$(1+x)^k(1+x)≥1+kx+x\Leftarrow$ by I.H.
$(1+kx)(1+x)≥1+kx+x\Leftarrow$
$kx^2+kx+x+1≥1+kx+x\Leftarrow$
$kx^2≥0$
which is true as $k \ge 1 > 0$.
This would be a correct induction step.
But my advice on a stylish and better induction step would be:
......
We are assuming that $(1+x)^k \ge 1 + kx$ and wish to show this implies $(1+x)^{k+1} \ge 1 + (k+1)x$.
$(1+x)^k \ge 1+kx$ (we are presuming this) and as $x > -1$ we know $1+x > 0$ and thus:
$(1+x)^k(1+x) \ge (1+kx)(1+x)$ so
$(1+ x)^{1+x} \ge 1 + kx + x + kx^2= 1+(k+1)x + kx^2$
And as $k$ and $x^2$ are non-negative, $kx^2 \ge 0$, so
$(1+x)^{1+x} \ge 1+(k+1)x + kx^2 \ge 1+(k+1)x$.
......
but ultimately the proof will need to be written by you.
......
I just realized no-one has yet addressed that claiming $(1+x)^k(1+x) \ge (1+kx)(1+x)$ requires knowing that $1+x > 0$.  This is why the condition $x > -1$ is required.
A: Theorem: If $x > -1$ then for all $n \geq 1$, $(1+x)^n \geq 1+nx$
Proof: By induction, the base case is $n=1$ where
$$1+x \geq 1+x$$
is true. For the induction hypothesis, we will assume that 
$$(1+x)^k \geq 1 + kx$$
for some $k\in\mathbb N$. Then, for the inductive step, we need to show that
$$(1+x)^{k+1} \geq 1 + (k+1)x$$
Starting from the LHS
\begin{align}(1+x)^{k+1}&=(1+x)^k(1+x)\\&\geq (1 + kx)(1+x)\\&=1 + kx + x + kx^2 \\&\geq1 + kx + x\\&=1 + (k+1)x\end{align}
Hence, we see that the theorem holds.
A: then would this be a better proof?:
$\forall x > -1$ we know that $kx^2 \geq 0$
by adding $kx+x+1$ to both sides:
$kx^2+kx+x+1 \geq kx+x+1$
$(1+kx)(1+x) \geq kx + 1 + x$
$(1+x)^k (1+x) \geq kx + 1 + x$ by I.H.
$(1+x)^{k+1} \geq  1 + (k+1)x$
Hence proved?
