Here is yet another answer complementing to already completely correct answers. As chi and Draconis state, the key property is $L^{m+n}=L^m L^n$. As Mickey states but does not explain why, this leads to $L^0=\{\epsilon\}$. Draconis' answer apparently makes it too easy to read as an "arbitrary" choice, or even that it is an artificial choice replacing some other, "more natural" choice.1
Here's a somewhat more general perspective. Virtually always when we use the notation $x^n$ where $n$ is a natural number, we intend $x^{(-)}$ to be a monoid homomorphism from the natural numbers equipped with the additive structure. This means $x$ itself must be an element of some specified monoid. Usually, this is implicitly specified by being the one we multiplicative notation for for elements of the relevant set. The unit and operation of the additive monoid on natural numbers are $0$ and $+$. For $x^{(-)}$ to be a monoid homomorphism means $x^0=u$ and $x^{m+n} = x^m \star x^n$ where $u$ and $\star$ are the unit and operation of the specified monoid on the set containing $x$.
Now the set of finite sequences(or lists) on an alphabet $\Sigma$, often written $\Sigma^*$, is actually the free monoid2 on $\Sigma$. The unit is the empty sequence $\epsilon$ and the operation is concatenation which I'll write as $\frown$ to have it be explicit. This leads to exactly the behavior you describe on the comment to the question. That is, for words $a$, $a^{m+n}=a^m\frown a^n$ and $a^0=\epsilon$. Next, given any monoid on a set $X$, we can define a new monoid on the powerset of $X$ (those elements are subsets of $X$). If $u$ and $\star$ are the unit and operation of a given monoid on $X$, then $\{u\}$ is the unit for the operation (which I'll call $\dot\star$), $S\ \dot\star\ T =\{s\star t\mid s\in S \land t \in T\}$. (Prove this.) This is, of course, exactly the monoid structure we're using when we talk about concatenating languages (i.e. subsets of $\Sigma^*$). This immediately leads to $L^0=\{\epsilon\}$ and $L^{m+n}=L^m\ \dot\frown\ L^n$.
There are many other monoid structures we can place on the power set of $X$. For example, for any set $X$ which need not be a monoid. We have the following two monoids on the power set of $X$: 1) unit $X$ and operation $\cap$, 2) unit $\varnothing$ and operation $\cup$. (Prove that these are monoids.) If you wanted $L_1 L_2$ to mean $L_1\cup L_2$, then we'd have $L^0 = \varnothing$ and $L^{m+n}=L^m\cup L^n$, but $L_1 L_2$ usually means $L_1\ \dot\frown\ L_2$ as above for which the unit, and thus $L^0$, is $\{\epsilon\}$. Choosing $L^0$ to be $\varnothing$ while still keeping $L_1 L_2$ as $L_1\ \dot\frown\ L_2$ means $L^{(-)}$ is no longer a monoid homomorphism. This will lead to all kinds of unexpected behavior, clumsy special-cases, and missed connections.
1 Draconis' point is more that definitions can't be wrong. One is free to define things however they like no matter how clumsy, awkward, and confusing the result may be. However, some definitions are better than others.
2 Formal language theory is very much a study of various monoids and operations upon monoids. It's pretty close to just being "Monoid Theory".