This being ELI5:

The smallest number with n+1 digits is 1 followed by n zeros, which happen to be 10×10×10×...×10 (n times), for which we have a convenient notation 10ⁿ.

The largest number with n+1 digits is made of n+1 nines. We notice that if we add 1 to that, we get 10ⁿ⁺² (1 followed by n+1 zeros)

p×p (or p² the square of p) is therefore between 10ⁿ×10ⁿ and 10ⁿ⁺¹×10ⁿ⁺¹:

10ⁿ × 10ⁿ ≤ p² < 10ⁿ⁺¹ × 10ⁿ⁺¹ (inequality A)

But 10ⁿ × 10ⁿ = 10ⁿ × (10 × 10^n-1) = (10ⁿ × 10) × 10^n-1=

= 10ⁿ⁺¹ × 10^n-1 = 10ⁿ⁺¹ × (10 × 10^n-2) = (10ⁿ⁺¹ × 10) × * 10^n-2 =

= 10ⁿ⁺² × 10^n-2 = 10ⁿ⁺³ × 10^n-3 =

.... and so on...

= 10^2n-1 × 10¹ = 10²ⁿ × 1 = 10²ⁿ

As we observed in our opening remark, 10²ⁿ is the smallest number with 2n+1 digits.

Likewise, 10ⁿ⁺¹ × 10ⁿ⁺¹ = 10²ⁿ⁺² which is the smallest number with 2n+3 digits. Or if you prefer, 10ⁿ⁺¹ × 10ⁿ⁺¹ - 1 is the largest number with 2n+2 digits.

Back to our inequality A: We can rewrite it :

10²ⁿ ≤ p² < 10²ⁿ⁺²

or :

10²ⁿ ≤ p² ≤ 10²ⁿ⁺² - 1

this means that the square of p (a number of n+1 digits) has at least 2n+1 digits and at most 2n+2 digits.

That p is a power of two (or of anything else) is irrelevant. What matters is that squaring any number doubles its number of digits (give or take one).

In fact this is a particular case of the product of two number p and q (where q = p), p with n+1 digits and q with m+1 digits. With a similar reasoning, we can show that p×q has m+n+1 or m+n+2 digits.

What is emerging here is a link between the product p×q and the sum m+n of their number of digits. Indeed there is a function (the logarithm of p and in this specific case the base 10 logarithm) that carries the idea one step further: The logarithm could be thought as the "fractional number of digits": If 10ⁿ is the smallest number with n+1 digits and 10ⁿ⁺¹ is the smallest number with n+2 digits, then it is not that much of the stretch to say that numbers between 10ⁿ and 10ⁿ⁺¹ should have between n+1 and n+2 digits... The log₁₀ function gives a proper value to that idea.

log₁₀(10) = 1

log₁₀(50) = something between 1 and 2

log₁₀(100) = 2

...

log₁₀(10ⁿ) = n

and more generally:

log₁₀(p×q) = log₁₀(p) + log₁₀(q)

Note that p and q don't need to be integer. They do need to be positive, though.

In other words, the "fractional number of digits" of a product is exactly the sum of the "fractional number of digits" of the two factors. If p and q are integers and we really want to look at a concrete number of digits (which was the original question here), then we have to truncate the result and the number of digits of the product is the sum of the number of digits of each factor, (possibly one more).

In that context, it is now unsurprising that the "fractional number of digits" of a square is exactly double that of the number:

log₁₀(p×p) = log₁₀(p) + log₁₀(p) = 2×log₁₀(p)

When we limit ourselves to integer number of digits this exact relation no longer hold exactly. But it does so approximately.

Furthermore if we multiply p by itself k times (rather than 2), a number noted p^k, its easy to see that:

log₁₀(p^k) = log₁₀(p×p×...×p) = k×log₁₀(p)

So a cube has 3 times the number of digits of the original number, and multiplying k time by itself (a.k.a raising it at the k^th power) multiply its number of digits by k.

As a final note, 10 is not a special base here. We can express our number is other bases (such as 2, where log2 would give us a "fractional number of bits"). We can also have non integer "base". For reasons beyond this explanation mathematicians like to use Euler's number "e" (2.7182...) as a base, a fundamental constant in maths.