$\require{enclose}\enclose{horizontalstrike}{\rm Greatest}\!$ Least prime factor of $n$ is less than square root of $n$, proof I remember reading this somewhere but I cannot locate the proof.
It is the smallest prime factor that is less than or equal to $\sqrt{n}$, unless $n$ is prime.  One proof is as follows: Suppose $n=ab$ and $a$ is the smallest prime factor of $n$, and $n$ is not prime.  Since $n$ is not prime, we have $b\ne1$.  Since $a$ is the smallest prime factor of $n$, we have $a\le b$.  If $a$ were bigger than $\sqrt{n}$, then $b$ would also be bigger than $\sqrt{n}$, so $ab$ would be bigger than $\sqrt{n}\cdot\sqrt{n}$.  But $ab=n$.
The answer is quite simple.

Given that n is composite so $n=p\cdot q\cdot r\cdot s\cdot t\cdots$ where $p,q\cdots$ are primes by fundamental theorem of arithmetic. 

Now assume without loss of generality that $p&lt;q&lt;r&lt;\cdots$. So $p&lt;q\cdot r\cdot t\cdots$. 

Now take square root on both sides then $\sqrt{p}&lt;\sqrt{q\cdot r\cdot s\cdot t\cdots}$. Now multiply both sides by $\sqrt{p}$. We get $p&lt;\sqrt{p\cdot q\cdot r \cdot s \cdot t\cdots}$ and $\sqrt{p\cdot q\cdot r\cdot s\cdot t\cdots}$ is equal to $\sqrt{n}$.

Hence established that $p \leq \sqrt{n}$.