How does one actually show from associativity that one can drop parentheses?

You can prove by induction on the number of factors that all such products are equal. We show that if you have $a_1,\ldots,a_n$, then any way to place the parenthesis will yield the same result as $a_1(a_2(\cdots(a_{n-1}a_n)\cdots))$.

If the number of factors is $1$ or $2$, then the result is immediate. If the number of factors is $3$, then it follows precisely by associativity: $(a_1a_2)a_3 = a_1(a_2a_3)$.

Assume the result holds for any $k$, $1\leq k\lt n$, and you have some product of $n$ elements $a_1,\ldots,a_n$, in order, with parentheses placed in some arbitrary (but valid) manner. Then there exists $k$, $1\leq k\lt n$, such that the product is of the form $$w_1(a_1,\ldots,a_k)\cdot w_2(a_{k+1},\ldots,a_n)$$ where $w_1(a_1,\ldots,a_k)$ is just an expression involving the product of $a_1,\ldots,a_k$ with parentheses placed in some fashion, and similarly with $w_2$. (We are just looking at the "last" product to be done; e.g., if you have $((a_1a_2)a_3)(a_4a_5)$, we would have $k=3$, $w_1(a_1,a_2,a_3) = (a_1a_2)a_3$, and $w_2(a_4,a_5) = a_4a_5$; if you have $a_1((a_2a_3)(a_4a_5))$, then $k=1$, $w_1(a_1) = a_1$, and $w_2(a_2,a_3,a_4,a_5) = (a_2a_3)(a_4a_5)$; etc).

By induction we can write $$w_1(a_1,\ldots,a_k) = a_1(a_2(\cdots(a_{k-1}a_k)\cdots))$$ so \begin{align*} w_1(a_1,\ldots,a_k)w_2(a_{k+1},\ldots,a_n) &= \Bigl(a_1\bigl(a_2(\cdots(a_{k-1}a_k)\cdots)\Bigr)\cdot w_2(a_{k+1},\ldots,a_n)\\\ &= a_1\cdot\Bigl(a_2(\cdots(a_{k-1}a_k)\cdots)w_2(a_{k+1},\ldots,a_n)\Bigr) \\\ &= a_1\cdot\Bigl( a_2(a_3(\cdots(a_{n-1}a_n)\cdots)\Bigr) \end{align*} where the last equality follows from the induction hypothesis applied to a product with $n-1$ factors, and the immediately preceding equality by simple associativity of three factors, applied to $a_1$, $a_2(\cdots(a_{k-1}a_k)\cdots)$, and $w_2(a_{k+1},\ldots,a_n)$. (Note: If $k=1$, then you can skip directly from the first line to the last line without having to invoke associativity; the argument still holds.)

Thus, any expression of a product of $a_1,\ldots,a_n$, in that order, with parentheses placed in an arbitrary but valid manner, is equal to $$a_1(a_2(\cdots(a_{n-1}a_n)\cdots)),$$ This completes the induction and the proof.

Since all ways of associating a product of $n$ factors, $a_1,\ldots,a_n$ yield results that are equal to one another, we may freely write any such product simply as $a_1\cdots a_n$ to represent their common value.

Hint $ $ It's very easy: keep pushing ')'s rightward using the rewrite rule $\rm\ (xy)z\ \to\ x(yz).\,$ This process terminates with the right-associated normal form where all ')'s are at the right end, namely $\rm\ a(b(c(d(\cdots))),\, $ written $\rm\ abc\cdots $ (unambiguous - since every bracketing rewrites to it by below).

Below is a proof sketch, by induction on the length $\rm\:\ell(Z) = $ number of operands (factors)

$\rm \ell(Z)\, =\, 1\!:\ \ \ Z = a\ $ is in normal form

$\rm \ell(Z)\, >\, 1\!:\ \ \ Z = XY\ $ where $\rm\:\ell(X),\ell(Y)< \ell(Z)\,$ splits into $2$ cases:

$\rm\ \ \ell(X)\! =\! 1\!:\ \ \ XY = aY = a(b(c(\cdots)))\ $ by induction applied to $\rm Y$

$\rm\ \ \ell(X)\! >\! 1\!:\ \ \ XY = \underbrace{(a \bar X)Y = a(\bar X Y)}_{\rm associative\ law} = a(b(c(\cdots)))\ $ by induction applied to $\rm X,\, $ then $\rm\bar XY$

The proof used only the associative law $\rm\, (XY)Z = X(YZ)\,$ so works for any associative operation.

That «one can drop the parentheses» really means that «no matter how you put the parentheses in, the result will be the same».

To prove such a statement, what one usually does is pick one specific way of putting in parentheses, and shows that any other way gives the same result as the one we picked. This is done by induction in the number of factors.

Some enterprising soul might post a complete proof...