Step in Understanding Mersenne Primes In this video https://www.youtube.com/watch?v=PYM-UYtLVDU the presenter claims that $2^6 -1$ = $111$ $111$ in base 2, and that this implies it is equal to $111 * something$ and is therefore not prime. What does he mean by this please? I get the conversion to binary, but I don't see why splitting the binary representation in this way implies compositeness.
In base 10 I guess the equivalent is to claim that $999$ $999$ is composite based on the fact that there is a repeating pattern of digits.
How is this step made please?
 A: Starting from the fact that a number whose binary representation is mabe only by $1$s is of the form $2^n-1$, you ask for some condition on $n$ in order to have $2^n-1$ prime.
This is not easy, and actually it is an open problem, anayway, there is an easy necessary (but unfortunately not sufficient) condition: $n$ must be prime. Indeed we can prove that $2^n-1$ is composite in $n$ is composite.
First, suppose $n$ even, so $n=2m$. Then $2^n-1=2^{2m}-1=(2^m)^2-1^2$ is a difference of square, and you can factorize it this way
$$(2^m)^2-1^2=(2^m-1)(2^m+1).$$
If $n$ is a composite odd, you can proceed in a similar way. Suppose $n=ab$, for some $a,b\in\mathbb{N}$ and $a,b>1$, then $2^n-1=2^{ab}-1=(2^a)^b-1^b$ is a difference of $b^\text{th}$ powers and so
$$
2^n-1=(2^a)^b-1^b=(2^a-1)((2^a)^{b-1}+(2^a)^{b-2}+\ldots+(2^a)+1).
$$

The same result can be proved also in a different way (which I think is more in the spirit of your original question). Let $n=ab$ as above, then we know that the binary representation of $2^{n}-1=2^{ab}-1$ is made by $ab$ consecutive $1$s. Now, perform the division of such number by $2^a-1$, which is a number whose binary representation is made by $a$ consecutive $1$s. We get
$$
\require{enclose}
\begin{array}{r}
                10\ldots01\ldots0\ldots01\  \\[-3pt]
\underbrace{1\ldots 11}_{a\text{ digits}} \enclose{longdiv}{\underbrace{\underbrace{1\ldots 11}_{a\text{ digits}} \underbrace{1\ldots 11}_{a\text{ digits}}\ldots\underbrace{1\ldots 11}_{a\text{ digits}}}_{b\text{ blocks}}} \\[-3pt]
     \underline{11\ldots 1}\phantom{\ldots00\ldots 00\ldots001} \\[-3pt]
                0\underline{1\ldots11}\phantom{\ldots00\ldots 01}  \\[-3pt]
     \ldots\phantom{0\ldots 0} 
\end{array}
$$
so the division is exact and the quotient is the number whose binary representation is a $1$ followed by $b-1$ blocks made of $a-1$ consecutive $0$s followed by a $1$.

Coming back to your original question, from my first observation, it follows that $2^6-1=63$ can be divided at least by $2^2-1=3$ and $2^3-1=7$, as $6=2\cdot 3$.
Form the second one, we have that $2^6-1$ in binary is $111\ 111$, so it can be divided by $111$ (i.e., by $7$), but we can also write $11\ 11\ 11$, so it can be divided by $11$ (i.e., by $3$).
Lastly, I want to add that $2^n-1$ can be a composite number also if $n$ is prime: the smallest counterexample is for $n=11$, as $2^{11}-1=2047=23\cdot 89$.
