I'd use a geometric argument combined with the principle of induction to convince your son. A simple, but fairly rigorous proof that is probably understandable to a middle schooler taking algebra.

Have your son imagine a square sheet of paper with side length $b$, where $b$ is very large, then by increasing its side length by $1$ we get a strip of paper added to the top and side of square, but increasing its *dimension* by $1$ will add far more material, since the strip of paper can clearly be cut and fit completely inside the cube with plenty of room to spare.

Therefore, $(b + 1)^2 < b^{2 + 1}$ (again, reminding your son that $b$ is very large).

In fact, if you like, you can just exhibit such a $b$ without giving the argument above, but I think it may help it sink in that increasing the variable in the exponential increases the volume of the square by an entirely new *dimension* while increasing the variable in a square just adds a tiny 2d slit of paper in comparison.

Now we just "repeat" the argument (you can use the word 'induction' if you like, but I think 'repeat' gets the idea across). Suppose that we know

$$(b + n)^2 < b^{2 + n}$$

for some $n$. We want to show that

$$(b + n + 1)^2 < b^{2 + n + 1}$$

we can rewrite this as

$$(b + n)^2 + 2(b + n) + 1 < b^{2 + n} b$$
The LHS rewrite can be seen geometrically. The side and top strips have length $(b + n)$ and height 1, and the corner has area 1, a diagram can be drawn to see this.

(Alternatively, not sure if they use this term in school anymore, but this way of rewriting the LHS was called "FOILing" when I was in middle school)

This inequality is definitely true, since

$$b^{2 + n} b = b^{2 + n} + \dots + b^{2 + n} \text{ ($b$ times)} $$,

This can also be argued geometrically, but it's trickier to visualize in dimensions higher than 3.

By our assumption, $$(b + n)^2 < b^{2 + n}$$

Therefore, $$(b + x)^2 < b^{2 + x}$$ for all natural numbers $x$ greater than or equal to 1.

If your son is interested in these types of things, then thinking about 'higher' dimensions will probably add an air of intrigue to the whole thing. The above induction step can be explained geometrically, and even still remain rigorous if a bit more work is put into it.

The idea is that, when you add 1 to the length of a square, you are adding only a 2 dimensional amount of stuff, but when you add 1 to the *dimension* of a square, you are adding a *3* dimensional amount of stuff.

When you add two to the length of the square, you are *still* adding only a 2 dimensional amount of stuff. But when you add two to the dimension of a square, not only are you adding the 3 dimensional amount of stuff, but you are also adding *4* dimensional amount of stuff now as well, and so on.