For example,
(6) The sequence of primes is endless.
For, if $p$ is any prime, the number ${\begin{array}{|c}\color{red}p\\\hline\end{array} + 1}$ is greater than $p$ and is not divisible by $p$ or by any smaller prime. If then ${\begin{array}{|c}p\\\hline\end{array} + 1}$ is not a prime, it must have a prime divisor greater than $p$, and in either case a prime greater than $p$ exists.
And here is another one:
9. Theorems.
(1) The product of any $n$ consecutive integers is divisible by $\begin{array}{|c}n\\\hline\end{array}$
For $(m+1)(m+2)...(m+n)/\begin{array}{|c}n\\\hline\end{array} = \begin{array}{|c}\color{red}{m + n}\\\hline\end{array}/\begin{array}{|c}\color{red}m\\\hline\end{array}\begin{array}{|c}\color{red}n\\\hline\end{array}$, and to show that the last expression is an integer it is sufficient to show that any prime $p$ which occurs in $\begin{array}{|c}\color{red}m\\\hline\end{array}\begin{array}{|c}\color{red}n\\\hline\end{array}$ occurs to at least as high a power in $\begin{array}{|c}\color{red}{m+n}\\\hline\end{array}$. Thus we have to show that
$$I[(m+n)/p]+I[(m+n)/p^2]+I[(m+n)/p^3]+... \\ \geq I[m/p]+I[m/p^2]+I[m/p^3]+... \\ +I[n/p] +I[n/p^2]+I[n/p^3]+...$$