How to prove that representations on $S^k(V), \bigwedge ^ k(V)$ are irreducible? Given a $\mathbb{C}$ vector space $V$, let $GL(V)$ act on $\bigotimes^k(V)$ via: 

$GL(V) \times \bigotimes^kV \to \bigotimes^k(V), \ (A,v_1\otimes...\otimes v_k)\mapsto Av_1\otimes...\otimes Av_k.$

I want to show that the representations on symmetric and exterior powers, $S^k(V), \bigwedge^k V$ are irreducible. 
I have tried the following: 
Consider the representation on $S^k(V).$ Let $\phi:GL(V)\to \mathbb{C}$ be the character corresponding to the representation on $S^k(V).$ Let $\phi_1,...,\phi_h$ be the irreducible characters with multiplicities $n_1,...,n_h$, and $W_1,..,W_h$ be the corresponding representations, i.e. $$S^k(V)=\bigoplus_{i=1}^h n_iW_i \ \mbox{ and } \ \phi=\sum_{i=1}^h n_i\phi_i, $$ where $n_i=(\phi|\phi_i).$ By orthogonality relation on characters, we have : $$(\phi|\phi)=\sum_{i=1}^h n_i ^2.$$ Now, I know from a theorem that $S^k(V) $ is irreducible iff $(\phi|\phi) =1.$ But I don't know how to show this sufficient condition. 
I am very new to representation theory. Please help!
 A: As in Nate's comment, the most direct way to do this is probably to consider the decompositions of these representations into common eigenspaces for the subgroup $T$ of diagonal matrices. Thus for each homomorphism $\lambda: T \rightarrow \mathbb{C}^\times$ and each representation $M$ of $G=GL_n(\mathbb{C})$ let
$$M_\lambda=\{m \in M \ | \ tm=\lambda(t)m \ \hbox{for all $t \in T$}\}.$$ The fundamental fact we shall use is that for a module $M$ that is equal to the direct sum
$$M=\bigoplus_\lambda M_\lambda$$ of $T$-weight spaces, every submodule $N$ is also such a direct sum. In fact, if $m=\sum m_\lambda$ with $m_\lambda \in M_\lambda$ and $m \in N$ then each $m_\lambda$ belongs to $N$ as well.
Assuming this fact, the strategy to prove that a given $M$ is irreducible is as follows: suppose $N$ is a non-zero submodule. Then it contains some $T$-eigenvector. Using this $T$-eigenvector, we try to produce all the other $T$-eigenvectors and obtain $N=M$. We will now carry this procedure out.
Let $e_1,\dots,e_n$ be the standard basis of $V=\mathbb{C}^n$, and let $\lambda_1,\dots,\lambda_n:T \rightarrow \mathbb{C}^\times$ be the characters of $T$ corresponding to its diagonal entries, with
$$t e_i=\lambda_i(t) e_i.$$ The $G$-module $\wedge^k V$ is $T$-diagonalizable with
$$t(e_{i_1} \wedge \cdots \wedge e_{i_k})=\lambda_{i_1}(t) \cdots \lambda_{i_k}(t)e_{i_1} \wedge \cdots \wedge e_{i_k}.  $$ This formula shows that the $T$-eigenspaces are all one dimensional, so that if $N$ is a non-zero submodule then it contains some decomposable wedge $e_{i_1} \wedge \cdots \wedge e_{i_k}$. Now given another decomposable wedge $e_{j_1} \wedge \cdots \wedge e_{j_k}$ we can choose a permutation matrix $g$ achieving $g(e_{i_l})=e_{j_l}$ for all $1 \leq l \leq k$. It follows that $N=M$. 
The proof for the symmetric power $S^k(V)$ is similar but slightly more complicated. Arguing as above using the fact that for the symmetric powers the $T$-eigenspaces are $1$-dimensional, a non-zero submodule $N$ must contain some monomial
$$e_1^{i_1} \cdots e_n^{i_n}.$$ Assuming $i_j \neq 0$ we apply a matrix $g$ with $g(e_j)=e_j+e_k$ for some $k$ and $g(e_i)=e_i$ for all $i \neq j$. Then expanding $g(e_1^{i_1} \cdots e_n^{i_n})$ as a sum of monomials, one of these is $$e_1^{i_1} \cdots e_j^{i_j-1} \cdots e_k^{i_k+1} \cdots e_n^{i_n}.$$ The fundamental fact with which I began this post shows that $N$ contains all monomials.
(At the last stage, the coefficient of $e_1^{i_1} \cdots e_j^{i_j-1} \cdots e_k^{i_k+1} \cdots e_n^{i_n}$ in the expansion is a binomial coefficient that may be divisble by $p$, so that in characteristic $p$ the argument breaks down precisely here. It seems that the exterior powers and sufficiently small symmetric powers are still irreducible, though.)
