Why aren't units with powers, like cm³, surrounded by parentheses? Since $\renewcommand{\unit}[1]{\,\mathrm{#1}} 1\unit{dm} = 10^{-1}\unit{m}$, it follows that $1\unit{dm^3} = 10^{-1} \times 10^{-1} \times 10^{-1} \unit{m^3} = 10^{-3} \unit{m^3}$.
However, in regular mathematics the following equation holds true:
$$a\,b^{3} = a\,b\,b\,b$$
By the above, the cube unit should expand as follows:
$$\mathrm{dm^3} = \mathrm{dmmm}$$
While in actual usage (as seen in the second equation) the expansion is $\mathrm{dddmmm}$, which would arise from using $\mathrm{(dm)^3}$ instead.
$$\mathrm{(dm)^3} = \mathrm{dddmmm}$$
So shortly: why aren't parentheses (commonly?) used in units?
 A: They're not used because it's ugly to read such texts with parentheses and it's time-consuming to write it down.
A decimeter is indeed a "product" of "deci" and a meter, so the origin is analogous to the product of two real numbers $ab$. But once we define the new derived unit ${\rm dm}$, we treat it as a single object, so it really means what you would call $({\rm dm})$.
Or if one wants to be really picky: in $ab$, the tiny space in between the two letters is a space in italics which may be interpreted as a multiplication of variables. However, the tiny space between ${\rm d}$ and ${\rm m}$ in "decimeter" is a Roman font minispace, and that isn't interpreted as a product anymore. That's why ${\rm dm}$ is interpreted as a whole.
A: $1\text{ dm}^3$ can be seen as a shorthand way to write $1\text{ dm}\cdot 1\text{ dm}\cdot1\text{ dm}$. The motivation behind this is that the quantity $1\text{ dm}^3$ represents the volume of a cube with each side of length $1\text{ dm}$, and the way to find the volume is to multiply the three sides: $LWH=(1\text{ dm})\cdot(1\text{ dm})\cdot(1\text{ dm})=1\text{ dm}^3$.
In other words, view "dm" as a new unit, not as a product. It might be less confusing to write $1\ (\text{dm})^3$ instead of $1\text{ dm}^3$, but that's the norm.
It may be beneficial to replace "dm" with some other measure that is represented by a single letter; this way you don't run into the issue of wanting to distribute the power. But it's probably better to get used to the norm.
As bdesham points out, it's important to note that some quantity like $4\ \text{dm}^3$ is not short for $4\ \text{dm}\cdot 4\ \text{dm}\cdot 4\ \text{dm}$, but rather $4\ \text{dm}= 4\cdot\left( 1\ \text{dm}\cdot 1\ \text{dm}\cdot 1\ \text{dm}\right)$. In other words, you have 4 cubes of side length $1\ \text{dm}$.
A: In your example, ab is implied to be a multiplication - a*b; but dm is a single indivisible token. 
Imagine second2 - that doesn't imply that the last letter should be squared, and it's the same with decimeters.
Mathematical notation is often ambiguous, and leaves many things implied and underspecified with the expection that the reader will fill the gaps properly - the assumption of ab = a*b is one of them.
A: The thing is that $\mathrm{dm}$ is a single symbol, not a combination of two symbols.
Yes, it can be understood in terms of a prefix and a base indicator, but it is still a single symbol. An analogy to the concatenation of variable is inappropriate.
Reference to an authoritative statement:

The grouping formed by a prefix symbol attached to a unit symbol constitutes a new inseparable unit symbol (forming a multiple or submultiple of the unit concerned) that can be raised to a positive or negative power and that can be combined with other unit symbols to form compound unit symbols.
Example: $\renewcommand{\unit}[1]{\,\mathrm{#1}} 2.3\unit{cm^3} = 2.3\unit{(cm)^3} = 2.3 \unit{(10^{–2}\,m)^3} = 2.3 \times 10^{–6} \unit{m^3}$

A: This is just a convention and nothing more. We treat the $\mathrm{dm}$ as a single symbol. Note that there are much stranger notations of powers, such as squared sines:
$$ \sin^2(x) = \big(\sin(x)\big)^2 $$
You can not treat this formally as written - it's not a square of the sine, but the square of the value it returned.
