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So:

$$ (1+x)^n ≥ 1 + nx $$

So he checks for 1, and get:

$$ 1+x ≥ 1+x $$

Next for variable k:

$$ (1+x)^k ≥ 1 + kx $$

Then the book wanna prove:

$$ (1+x)^{k+1} ≥ 1 + (k + 1)x $$

And here is books proof:

$$ (1+x)^{k+1} = (1+x)^k (1+x) ≥ (1+kx)(1+x) $$ $$ = 1+(k+1)x + kx^2 ≥ 1 + (k + 1)x $$

Finished! Well... How did the book get this: $(1+kx)(1+x)$ in that last part? Sorry, I'm so confused. Sorry if this to easy to be here. Thanks for all help helping me understand it!

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2 Answers

Your question is a great one, it is most welcome here! The answer is that, in a proof by induction, we first check the base case (here, it is $n=1$), and then, assuming the result is true for $n=k$, we prove that the result must also be true for $n=k+1$. In other words, we want to prove that $$\text{true for }n=k\implies\text{true for }n=k+1$$

Intuitively, this lets us say $$\begin{align} (\text{base case}) \qquad\qquad\qquad\qquad\qquad\qquad&\text{true for }n=1\qquad\checkmark\\ {\text{true for }n=1,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=2\qquad\checkmark\\ {\text{true for }n=2,\text{ and }\atop (\text{true for }n=k\implies\text{ true for }n=k+1)}\bigg\}\implies&\text{true for }n=3\qquad\checkmark\\ \vdots\end{align}$$

Thus, when we try to prove that the statement is true for $n=k+1$, i.e. $$(1+x)^{k+1} ≥ 1 + (k + 1)x,$$ we can use the assumption that the statement is true for $n=k$, i.e. $$(1+x)^k ≥ 1 + kx.$$ The reason why we have $$(1+x)^k (1+x) ≥ (1+kx)(1+x)$$ is that we are assuming $$(1+x)^k ≥ 1 + kx$$ is true, and then we multiply both sides by $(1+x)$.

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Actually, $n=0$ can be a base case as well :-) –  Asaf Karagila Aug 21 '11 at 14:44
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In your next to last line, you have $(1+x)^k,$ which you have assumed two lines above is greater than or equal to $1+kx$. So it made the substitution, using a $\ge$ sign. You might look at this answer, which has a detailed explanation of induction.

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