You can't prove it, since it is false. Take any finite-dimensional represention $\pi$ of $\mathfrak{sl}(2)$. And now consider the representaion $\pi^\star$ that it induces on $V\oplus V$. Then, whatever the dimension $d$ of the eigenspace of the highest weight $\Lambda$ of $\pi$ is, the dimension of the eigenspace of the highest weight $\Lambda$ of $\pi^\star$ is $2d$, and therefore it cannot possibly be equal to $1$.
As Tobias Kildetoft metioned, you can find details and proofs in Mazorchuk's "Lectures on sl2(C)-modules", section 3.4.
Your $P(\lambda)$ is a very good guess. But actually you need one more parameter to cover all weight modules, since you can vary the set of weights, as well as the central character.
Take $\xi \in \mathbb{C} / 2 \mathbb{Z}$ and $\tau \in \mathbb{C}$, and define the module $V(\xi,\tau)$ on basis $\{v_\mu \colon \mu \in \xi\}$ as follows:
- $h \cdot v_\mu = \mu v_{\mu}$,
- $f \cdot v_\mu = v_{\mu-2}$,
- $e \cdot v_\mu = \frac{1}{4} (\tau - (\mu+1)^2) v_{\mu+2}$.
These modules are sometimes called the principal series, and you can find all finite-dimensionals, Vermas, duals Vermas, as subquotients of them.
Anyway, the module $V(\xi,\tau)$ is simple if and only if $\tau \neq (\mu+1)^2$ for all $\mu \in \xi$. And such are pairwise non-isomorphic.
So, to your list you can add:
- $V(\xi,\tau)$, for $\xi \in \mathbb{C} / 2 \mathbb{Z}$ and $\tau \in \mathbb{C}$ such that $\tau \neq (\mu+1)^2$ for all $\mu \in \xi$.
And you get a complete list of mutually non-isomorphic simple weight $\mathfrak{sl}_2(\mathbb{C})$-modules.
Best Answer
Edit: Sorry, in an earlier answer I misread your post.
This is indeed true, assuming that the fundamental weights $w_1, ..., w_{n-1}$ are ordered according to the standard ordering of the simple roots in the Dynkin diagram. Namely, in general, the lowest weight of a highest weight representation $L(\Lambda)$ is the negative of the highest weight of the dual representation $L(\Lambda)^*$; and this, in turn, is given by $-w_0(\Lambda)$ where $w_0$ is the longest element of the Weyl group.
See Relationship between highest and lowest weights for a simple complex Lie algebra, finding highest weight of dual of a representation of a semisimple lie algebra, Highest weight of dual representation of $\mathfrak{sl}_3$, Irreducible Dual Representation.
The upshot is: The lowest weight of $L(\Lambda)$ is $w_0(\Lambda)$. Now in a root system of type $A_{n-1}$, $w_0$ is given on simple roots in the usual ordering by $w_0(\alpha_i) = -\alpha_{n-i-1}$, and is easily seen to do the same on the fundamental weights, i.e. $w_0(w_i) =-w_{n-1-i}$. Since its action is linear, this explains your formula.