Bitwise bit shifting and why this solution works - arrays

Question

I've been doing code fights on codefights.com and I came upon this problem below. I have figured out the problem on my own, but when i researched other peoples solutions I found one that was much shorter than mine, but I cant seem to understand why they did what they did.
The question goes:
You are given an array of up to four non-negative integers, each less than 256.
Your task is to pack these integers into one number M in the following way:
The first element of the array occupies the first 8 bits of M;
The second element occupies next 8 bits, and so on.
Return the obtained integer M.
Note: the phrase "first bits of M" refers to the least significant bits of M - the right-most bits of an integer. For further clarification see the following example.
Example
For a = [24, 85, 0], the output should be
arrayPacking(a) = 21784.
An array [24, 85, 0] looks like [00011000, 01010101, 00000000] in binary.
After packing these into one number we get 00000000 01010101 00011000 (spaces are placed for convenience), which equals to 21784.
Their answer was:
func arrayPacking(a []int) (sum int) {
for i := range a {
sum += a[len(a) - i - 1] << uint((len(a) - i - 1) * 8)
}
return
}
How is this code returning the right amount of shift just by using 0, 8, 16, etc. intervals? I've been researching bitwise a lot lately, but I still can't seem to reason why this works.

Answer 1

First, write the solution in Go. We convert little-endian, base-256 digits to a base-2 (binary) number. Shifting left by 8 bits multiplies by 256.
package main
import (
"fmt"
)
func pack(digits []int) (number int) {
// digits are little-endian, base-256 (1 << 8 = 256)
for i, digit := range digits {
number += digit << uint(i*8)
}
return number
}
func main() {
a := []int{24, 85, 0}
m := pack(a)
fmt.Println(m)
}
Playground: https://play.golang.org/p/oo_n7CiAHwG
Output:
21784
Now you should be able to figure out their ugly looking answer:
func arrayPacking(a []int) (sum int) {
for i := range a {
sum += a[len(a) - i - 1] << uint((len(a) - i - 1) * 8)
}
return
}

Answer 2

the bitshifting by multiples of 8 is the same as muliplying by multiples of 256, e.g. x << 0*8 == x * 256⁰, x << 1*8 == x * 256¹, x << 2*8 == x * 256² etc., so the code can be rewritten like this, using math.Pow:
func arrayPacking(a []int) (sum int) {
for i := range a {
sum += a[len(a) - i - 1] * int(math.Pow(256, (len(a) - i - 1)))
}
return
}
Or is your question why this sort of packing works?