Symmetric powers and alternating powers are irreducible modules over the general linear Lie algebra $\mathfrak{gl}(n)$ Let $n\geq 0$, and let $V\cong \mathbb{C}^n$ be the standard representation for $\mathfrak{gl}(n)$. Is there any proof for irreducibility for the symmetric power $S^{\ell}V$ and alternating power $\Lambda^{\ell}V$, for all $\ell \geq 1$ without using Weyl character formulas? Thanks! 

Fix a basis $(x_1,\dots,x_n)$ of $V$. A basis of $S^k V$ is given by the monomials $m_n=\prod x_i^{n_i}$ where $n=(n_i)$ ranges over $n$-uples with $\sum n_i=k$. The $m_n$ have distinct weights (*) for the action of the group of diagonal matrices. In particular, every invariant subspace $W$ has a basis made up of such monomials, and whenever some element $W$ has a nonzero coordinate in some monomial, then this monomial belongs to $W$.
Now start with a monomial $m_n$ in $W$. If $n$ is not supported by a single element, applying a suitable unipotent element, we can increase the value of some element. Iterating, we see that $W$ contains a monomial of the form $x_i^k$, and using permutations matrices, it contains all such monomials.
Again using unipotent elements, we see it contains all $x_i^\ell x_j^{k-\ell}$, and so on, we eventually get all monomials. (Argue by induction on $n$: then we have all monomials with $n_n=0$; then we get $m=\prod_{i<n}x_i^{n_i}x_n^s$ by induction on $s$: apply a unipotent matrix to the monomial $mx_1/x_n$).
The case of $\Lambda^k V$ starts similarly and finishes even more easily: $W$ is generated by monomials (with the restriction $n_i\le 1$ now). Then given a single monomial, we get all others using permutation matrices.  

(*) If you're not familiar with weights, here's a way to avoid it (even if it's the same principle). Choose a an algebraically free family $(t_1,\dots,t_n)$ and consider the diagonal matrix $d$ with diagonal $(t_1,t_2,\dots,t_n)$. Its action on $S^k V$ can be diagonalized and $\prod x_i^{n_i}$ is an eigenvector for the eigenvalue $\prod_{i=1}^nt_i^{n_i}$. The freeness assumption implies that these eigenvalues are pairwise distinct (actually is would have been enough to assume that the $t_i$ are multiplicatively a $\mathbf{Z}$-linear free family, so picking $t_i$ to be the $i$-th prime number works too). The argument also works for $\Lambda^k V$ since then the eigenvalues are the same, with the restriction $n_i\le 1$. Since $d$ is diagonal with eigenspaces of dimension 1, any $d$-invariant subspace is a sum of some of those eigenspaces.
