**Theorem** If $x>-1$ then for every $n \in \mathbb{N}$, $(1+x)^{n} \geq 1+nx$ **Proof** Base case: if $n=1$ then $(1+x)^{n}=1+x=1+nx$ so the statement is true for $n=1$. Now suppose true for $n=k$ for some $k \in \mathbb{N}$. So $(1+x)^{k} \geq 1+kx$. Note that $1+x>0$ so multiplying both sides of the inequality by it gives: $\begin{align} (1+x)^{k+1} &= (1+x)(1+x)^{k} \\ &\geq (1+x)(1+kx) \\ &= 1+x+kx+kx^{2} \\ &> 1+(k+1)x \end{align}$since $kx^{2}>0. \quad \square$ Common application is [[x=1+h proof]]. ### Theorems - [[Euler's Number as a Convergent Real Sequence]]. - [[There is a unique real number that is the square root of 2]] and [[Existence & uniqueness nth root of positive reals]].