Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality

In the setting of Commutative Rings, Fields and Ordered Fields: Standard Notation, let $R$, with $+$, $\cdot$, $0$ and $1$, be a commutative ring, and let $x,y\in R$ and $m,n\in\mathbb{N}_{0}$.

¶ $x^{m+n}=x^{m}x^{n}$ and $(x^{m})^{n}=x^{mn}$.

¶ $(xy)^{n}=x^{n}y^{n}$, $x^{1}=x$ and $1^{n}=1$; if $n\ge1$, then $0^{n}=0$.

¶ If $n\ge1$, then $x^{n}-y^{n}=(x-y)\sum_{k=1}^{n}x^{k-1}y^{n-k}$.

Let now $F$ be a field; let $x\in F$ and $n\in\mathbb{N}_{0}$.

¶ If $x\neq0$, then $x^{n}\neq0$ and $(x^{-1})^{n}=(x^{n})^{-1}$.

¶ If $x\neq1$ and $n\ge1$, then $\sum_{k=1}^{n}x^{k-1}=\dfrac{1-x^{n}}{1-x}$.

Let now $F$ be an ordered field; let $x,y\in F$ and $n\in\mathbb{N}_{0}$.

¶ If $x\ge0$, then $x^{n}\ge0$; if $x>0$, then $x^{n}>0$; and $|x^{n}|=|x|^{n}$.

¶ If $n\ge1$ and $0\le x<y$, then $x^{n}<y^{n}$.

¶ If $n\ge1$ and $x,y\ge0$, then $x\le y$ if and only if $x^{n}\le y^{n}$, and $x=y$ if and only if $x^{n}=y^{n}$.

¶ If $0\le x\le1$, then $0\le x^{n+1}\le x^{n}\le1$.

¶ If $x\ge1$, then $1\le x^{n}\le x^{n+1}$.

¶ Let $m\in\mathbb{N}_{0}$ with $m\le n$. If $x\ge1$, then $x^{m}\le x^{n}$; if $0\le x\le1$, then $x^{n}\le x^{m}$; and if $x>1$ and $m<n$, then $x^{m}<x^{n}$.

¶ If $x\ge-1$, then $(1+x)^{n}\ge1+n\,x$.