My son's Sum of Some is beautiful! But what is the proof or explanation?

Factor out the $2^n$ and you get: $2^n (100+20+8) = 2^n 128 = 2^{n+7}$ since $2^7 = 128$

It works because the number

$$ 128 $$ has two special properties: it is a power of $2$, and every digit is a power of $2$. This means that we can write it in two separate ways:

\begin{align} 128&=2^7\\ 128&=100\times2^0+10\times2^1+1\times2^3 \end{align}

Multiplying both sides by $2^n$ then gives:

$$ 2^{n+7}=(100\times2^n)+(10\times2^{n+1})+2^{n+3} $$

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Edit: others have done a better job of generalizing this, but I feel I should point out another obvious related sequence.

$128$ is not the only power of $2$ whose digits are all powers of $2$ (or $0$). For instance:

$$ 1024 = 2^{10} $$

So another, similar kind of relation is given by:

$$ 2^{n+10} = (1000\times2^n)+(10\times2^{n+1})+2^{n+2} $$

In your son's notation, that becomes:

Every one of these numbers is two times the number before it.
$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
I pick any one of them, times $1000$. Then I add the next one, times $10$. Then I add the next one.
If I then skip seven (!) ones and read the eighth, that one equals my sum!

See if he likes that one.

Here is another formula similar to your son's, using powers of $5$:

$$1000000\cdot5^n+100000\cdot5^{n+1}+10000\cdot5^{n+2}+1000\cdot5^{n+3}+100\cdot5^{n+4}+5^{n+6}=5^{n+9}$$

We can derive identities of a similar form more generally. Consider any $m$th degree polynomial $f(x)\in\mathbb{Z}[x]$ of the following form

$$f(x)=a_0+a_1x+\ldots+a_{m-1}x^{m-1}-x^m$$

Notice that all rational roots of $f(x)$ are integers. Suppose $f$ has a rational root $k$. Then we have the following equality:

$$\begin{align}k^{n+m}&=k^nk^m\\&=k^n(a_0+a_1k+a_2k^2+\ldots+a_{m-1}k^{m-1})\\&=a_0k^n+a_1k^{n+1}+a_2k^{n+2}+\ldots+a_{m-1}k^{n+m-1}\end{align}$$

Your son's formula involves the $7$th degree polynomial $f(x)=100+10x+x^3-x^7$ and the root $k=2$. Above I've used the polynomial $f(x)=10^6+10^5x+10^4x^2+10^3x^3+10^2x^4+x^6-x^9$ and the root $k=5$.

Here are other examples for $k=3$. Consider the polynomials $f(x)=3+8x+80x^3-x^7,$ $g(x)=2130+10x+x^3-x^7$, and $h(x)=99+687x+x^3-x^7$, and notice that $f(3)=g(3)=h(3)=0$. These give us the following identities:

$$3\cdot3^n+8\cdot 3^{n+1}+80\cdot3^{n+3}=3^{n+7}\\2130\cdot3^n+10\cdot3^{n+1}+3^{n+3}=3^{n+7}\\99\cdot3^n+687\cdot3^{n+1}+3^{n+3}=3^{n+7}$$

Try using this method to find identities for other values of $k$ (including negative integers!).