Something interesting that I found about some numbers - and would like to see if it's known

"Adding the digits until [you] get a number from $1$ to $10$" is the same as finding the remainder when dividing by $9$ (except that you would get $9$ instead of $0$ by adding digits, and there is no reason to stop with $10$: you can add the digits to get $1$).

It turns out, by something called Euler's Theorem, that if $a$ is not divisible by $3$, then $a^6$ always leaves a remainder of $1$ when divided by $9$. In particular, $2^6$ has remainder $1$ when divided by $7$.

Another property of remainders is that if $a$ leaves a remainder of $r$ when divided by $9$, and $b$ leaves a remainder of $s$, then $ab$ leaves the same remainder as $rs$.

So, that means that if $2^n$ leaves a remainder of $r$, then $2^n\times 2^6 = 2^{n+6}$ will leave the same remainder as $r\times 1 = r$; that is, the same remainder as $2^n$. So adding the digits of $2^n$ until you get a single number between $1$ and $9$ will give you the same answer as doing it for $2^{n+6}$. This is what you observed: $8=2^3$, and $512=2^9 = 2^{3+6}$. You will get the same answer ($7$) with $2^{15}$, $2^{21}$, $2^{27}$, etc.

I note that you were slightly off in describing $512$ as being "seven steps after" $8$: it's really only six steps later: $$8\stackrel{1}{\mapsto} 16 \stackrel{2}{\mapsto}32 \stackrel{3}{\mapsto}64\stackrel{4}{\mapsto}128\stackrel{5}{\mapsto}256\stackrel{6}{\mapsto}512.$$

This is indeed the case.

The value you get out at the end of all your summations is just the value of your number mod 9 (this is because any natural number is congruent to the sum of its digits mod 9). For example, $1024=9\cdot113+7$ is congruent to 7 mod 9. Also note that the numbers you're getting out by your procedure of successive doubling are just the powers of 2. Finally, I think you're off by one in your counting of sevens -- the powers of two that are working for you are $2^4=16$, $2^{10}=1024$, $2^{16}=65536$, etc.

So your fact is true because it's true for $2^4$, and each subsequent term differs by a factor of $2^6$ which is congruent to 1 mod 9 (either by direct computation or by Euler's Theorem -- $\phi(9)=6$).

So in general, the sum of the sum of the .... sum of the digits of your values $2^{4+6k}$ is given by $$ 2^{4+6k}\equiv 2^4\cdot (2^6)^k\equiv 7\cdot 1^k\equiv \boxed{7}\pmod{9}. $$

There is indeed a period of $6$. The sums of digits cycles through the numbers $1,2,4,8,7,5$.

You are doubling the number each time, but then the sum of the digits of that number will double as well, since it is a linear function of the number. This then shows that if you start with $1$ you will go through the cycle

$$1 \to 2 \to 4 \to 8 \to 16=7 \to 14=5 \to 10=1 \to \ldots$$