Why is the representation of $\mathfrak{sl}(2,\mathbb C)$ defined in this way irreducible? Let $\{h=\begin{bmatrix}
1 & 0 \\
0 & -1 
\end{bmatrix},e = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix},f = \begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}\}$ be the canonical basis of $\mathfrak{sl}(2,\mathbb C)$ and consider the linear representation $$\pi:\mathfrak{sl}(2,\mathbb C) \to \text{End}(V),$$
where $V=\text{Span}\{v_0,\dots, v_{n+1}\}$ (a $n+1$ dimensional vector space), defined on the basis $\{h,e,f\}$ by
$$\pi(h)v_i=(n-2i)v_i,$$ $$\pi(e)v_i=i(n-i+1)v_{i-1},$$ $$\pi(f)v_i=v_{i+1},$$
for $i=0,1,\dots,n$ with the convention that $v_{-1}=v_{n+1}=0$. It is not hard to check that $\pi$ preserves the Lie brackets.
Now, I would like to show $\pi$ defined in this way is irreducible from definition. Let $U\subset V$ be a nonzero $\pi$-invariant subspace. If I could prove that at least one of the vectors $\{v_0,\dots,v_n \}$ is in $U$, then I can use the relations $\pi(e)v_i=i(n-i+1)v_{i-1},\pi(f)v_i=v_{i+1}$ to conclude that all of $\{v_0,\dots,v_n\}$ are in $U$, which forces $U=V$.
But how do we prove that $U$ contains at least one of $\{v_0,\dots,v_n \}$? I guess we have to use $\pi(h)v_i=(n-2i)v_i$ in some way?

This is from Lie groups: Beyond an introduction by Knapp, page 64. It seems that the author does not assume any advanced theory for this verification. The notion of weight (vectors) has not been introduced at this point.
 A: Maybe the following adaptation/recapitulation of some basic results suits:
Suppose $W$ is a proper subrepresentation. Let $v=c_1v_1+\ldots+c_nv_n$ be a shortest (non-zero) linear combination of weight vectors $v_i$ lying inside $W$. Let $i_o$ be the lowest index such that $c_{i_o}\not=0$. Then (EDIT) unless already $\pi(h)(v)=(n-2i_o)v$, in which case $v=v_{i_o}$, the difference $v'=\pi(h)(v)-(n-2i_o)v$ is non-zero, is still in $W$, and is a shorter linear combination of weight vectors. (EDIT2) In formulas:
$$
\pi(h)(v) - (n-2i_o)v \;=\; \sum_i c_i(n-2i)v_i - \sum_i c_i(n-2i_o)v_i
$$
$$
\;=\; \sum_{i\not=i_o} c_i((n-2i)-(n-2i_o))v_i
\;=\; \sum_{i\not=i_o} (2i_o-2i) v_i
$$
Thus, no $v_i$ not already occurring in the linear combination occurs in the image, and $v_{i_o}$ no longer occurs.
Then since the formulas show that any one of the weight vectors generate the whole thing, we're done.
A: Here is yet another argument, which is fairly direct and which is repeated numerous times, in essentially this form, throughout representation theory. The fact from linear algebra that is crucial is this: if you have a linear operator $T$ acting on a $k$-dimensional vector space $V$ and the eigenvalues for $T$ are all distinct, with eigenvectors $v_1,\dots,v_k$, then any eigenvector for $T$ is a non-zero multiple of one of the $v_i$'s.
If you suppose $U$ is a non-zero $\pi$-invariant subspace, then since $\mathbf{C}$ is algebraically closed (this is not crucial, but is quick), $U$ contains an eigenvector for the operator $\pi(h)$. Since $\pi(h)$ acts diagonalizable on $V$ with eigenvectors $v_0,v_1,\dots,v_{n+1}$ of distinct eigenvalues $n,n-2,\dots,-n+2,-n$, any eigenvector for $\pi(h)$ in $V$ must be a non-zero multiple of some $v_i$. Thus $U$ contains some $v_i$, and the formulas for $\pi(e)$ and $\pi(f)$ show that $U$ contains all of $v_0,\dots,v_{n+1}$ and is hence equal to $V$.
