Why use $(1+p)^n\geq 1+np$ to prove that the successive powers of a number $q^n$ with $-1I'm reading Courant's What is Mathematics?
In page $64$, he gives the example of limit of the successive powers of $q$. If $-1<q<1$ then the successive powers of $q$ will approach zero as $n$ increases. Then he suggests that to give a rigorous proof of that assertion, one needs to use the inequality proved on page $15$, which is:
$$(1+p)^n\geq 1+np$$
But I'm not sure on how these things are connected - And I guess I'm unable to provide more information on my doubt, I just want to understand why that inequality is used to prove that that limit approaches $0$ as $n\rightarrow \infty$ for $-1<q<1$. 

I have something to add to this question. I guess I'm half the way to understand , but this is not the point. Reading the sentence again:

If $-1<q<1$ then the successive powers of $q$ will approach zero as $n$ increases.

I have an argument that could work as a proof, and it's a lot simpler: Suppose we take the number $10$, observation the successive powers of 10:
$$
\begin{array}{cr}
 10 & 10 \\
 10^2 & 100 \\
 10^3 & 1000 \\
 10^4 & 10000 \\
\end{array}$$
As we raise $n$ in $10^n$, we add one zero and the one goes on sliping to the left, this will happen for any number bigger than $1$ and different of $0$, the process will only take a little more time to happen, for $2^n$ we have:
$$
\begin{array}{cc}
 2 & 2 \\
 2^2 & 4 \\
 2^3 & 8 \\
 2^4 & 16 \\
\end{array}$$
If $a$ is smaller than $1$ in $a^n$, then we'll have for example:
$$\begin{array}{cc}
 \frac{1}{10} & 0.1 \\
 \left(\frac{1}{10}\right)^2 & 0.01 \\
 \left(\frac{1}{10}\right)^3 & 0.001 \\
 \left(\frac{1}{10}\right)^4 & 0.0001 \\
\end{array}$$
Then if the numbers are different of $1$ and $0$, I know that these numbers are going to walk. If $n<1$, it's going to walk to the right, and if $n>1$, the number will walk to the left. This looks pretty convincing to me so, why do I need to use Bernoulli's inequality? The answers given by John and RobJohn are useful, they tell me how to do it. But it's not really clear why I should use it instead of the method I just pointed.
 A: Let $|q| <1$, $q\neq 0$. Let $\frac{1}{|q|} =1+ b$ for some positive $b$. Then
$\frac{1}{|q|^n} = (1+b)^n \geq 1 + bn$. Which implies
$|q^n| \leq \frac{1}{1+nb}$. As we know the right hand side goes to zero, so is the left hand side. 
A: One way to show that $q^n\to0$ when $|q|\lt1$ is to look at logarithms. That is, $\log|q|\lt0$, so
$$
\begin{align}
\log\left(\lim_{n\to\infty}|q|^n\right)
&=\lim_{n\to\infty}n\log(q)\\
&=-\infty
\end{align}
$$
which essentially says that
$$
\lim_{n\to\infty}|q|^n=0
$$

However, Bernoulli's Inequality can be proven without using logarithms, and we can use that instead.
Suppose that $0\lt q\lt1$, then $\frac1q\gt1\iff r=\frac1q-1\gt0$. Then Bernoulli says that
$$
\begin{align}
q^n
&=\frac1{(1+r)^n}\\
&\le\frac1{1+nr}
\end{align}
$$
Since $r\gt0$, the Archimedean Property of the reals says that there is a $k$ so that $kr\gt1$. Then
$$
\begin{align}
q^{nk}
&\le\frac1{1+nkr}\\
&\le\frac1n
\end{align}
$$
ans therefore, we can deduce that
$$
\lim_{n\to\infty}q^n=0
$$
This maintains the spirit of the logarithmic proof without using logarithms.
