Equality of Bernoulli's inequality The simple form of Bernoulli's inequality is:
$$ (1+x)^n \geq 1+xn \quad \text{ where } n\in\mathbb{N} \wedge \ x \geq -1 $$
It is really easy to prove it via mathematical induction.
The only two options in which inequality becomes an equality are:


*

*$x=0$

*Any $x$ and $n = 1$
Now I wonder how do I prove that these two are the only options?
It would be better to prove that without using derivatives and so on...

Here is what I have tried:
Lets assume there is a special $t\neq 0$, $t\geq -1$ such as:
$$ (1+t)^n = 1+nt $$
Expanding the left part of an equation:
$$ (1+t)^n = 1 + nt + \binom{n}{2} t^2 + \ldots $$
This sum is obviously larger then $1+nt$ when $t>0$. So we have a contradiction.
How to get a contradiction when $-1 \leqslant t < 0 $?
 A: We can prove this via induction.
Let us fix $t \in [-1, 0)$. We show that if $n \ge 2$, then $(1 + t)^n > 1 + nt$.
Proof. Let us show the base case for $n = 2$.
$(1 + t)^2 = 1 + 2t + t^2 > 1 + 2t$, where the last inequality follows from the fact that $t^2 > 0$ as $t \neq 0$.
Now, let us assume that the statement is true for $n = k \ge 2$. We then have:
$$(1+t)^{k+1} = (1+t)^k(1+t) \ge (1 + kt)(1 + t).$$
This $\ge$ is because $1 + t \ge 0$ and $(1 + t)^k > 1 + kt$. (Note that we need not have strict inequality here.)
Now,
$$(1 + kt)(1 + t) = 1 + (k+1)t + kt^2 > 1 + (k+1)t.$$
Here, we have strict inequality as $k > 0$ and $t^2 > 0$.
Thus, we get that $(1 + t)^{k+1} > 1 + (k+1)t$.
By principle of mathematical induction, we are done.
A: For any $n \geq 2$ we assume that $(1+t)^n = 1+nt$:
$$ 1+nt=(1+t)^n=(1+t)^{n-1}(1+t) \geq (1+(n-1)t)(1+t)=1+nt+(n-1)t^2 \\[15px] 1 + nt \geq 1 + nt + (n-1)t^2 \\[15px] (n-1)t^2 \leq 0 $$
We get contradiction because this multiplication can be only positive ($t^2 > 0$, $n-1 > 0$). 
