Irreducible Module of $sl_{l+1}(\mathbb{C})$ as a Direct Summand

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This question is in relation to my previously asked question here Construction of an Irreducible Module as a Direct Summand.

Let $V_0$ be any arbitrary finite dimensional $sl_{\ell +1}(\mathbb{C})$ module and $V$ be the standard $(\ell+1)$ dimensional representation of $sl_{\ell +1}(\mathbb{C})$ . Then does there exist $m \in$ $\mathbb{N}$ such that $V_0$ occurs as an irreducible summand of $V^{\otimes{m}}$(the decomposition is upto isomorphism of $sl_{\ell+1}(\mathbb{C})$ modules)?

I have tried to solve this problem. Can anybody please verify if the solution is ok? Thanks for any help.

We know that $V_0$ is isomorphic to a $sl_{\ell+1}(\mathbb{C})$ module of highest weight $\lambda$, say $V(\lambda)$ where $\lambda$ is a dominant integral weight. Now if $\omega_1$,.....,$\omega_\ell$ denote the fundamental weights of the root system $A_\ell$, then we know that $\omega_i$ = $\epsilon_1$ + ....+$\epsilon_i$ where $i$ = $1,....,\ell$ and the highest weight of $(\wedge^1(V))^{\otimes{k_1}}$ $\otimes$ ......$\otimes$ $(\wedge^\ell(V))^{\otimes{k_\ell}}$ is $\lambda$ = $k_1\omega_1$ + $k_2\omega_2$ +....+$k_{\ell-1}\omega_{\ell-1}$ + $k_\ell\omega_\ell$, $k_i$ $\in$ $\mathbb{N}$ $\cup${0}.So we can define the projection map onto the wedge product given by $\phi$ : $V^{\otimes{m}}$ $\longrightarrow$ $(\wedge^1(V))^{\otimes{k_1}}$ $\otimes$ ......$\otimes$ $(\wedge^\ell(V))^{\otimes{k_\ell}}$ where $m$ = $k_1$ + $2k_2$ +........+$(\ell-1)k_{\ell-1}$ +$\ell k_{\ell}$. This is clearly a homomorphism of $sl_{\ell+1}(\mathbb{C})$ module and onto. So, we have $V^{\otimes{m}}$ $\simeq$ $ker\phi$ $\oplus$ $(\wedge^1(V))^{\otimes{k_1}}$ $\otimes$ ......$\otimes$ $(\wedge^\ell(V))^{\otimes{k_\ell}}$ by complete reducibility ($\because$ $sl_{\ell+1}(\mathbb{C})$ is semisimple) where we know that $V(\lambda)$ occurs as a direct summand in $(\wedge^1(V))^{\otimes{k_1}}$ $\otimes$ ......$\otimes$ $(\wedge^\ell(V))^{\otimes{k_\ell}}$. Thus we have our required $m\in$ $\mathbb{N}$. If anyone has got some other proof or suggestions or if this can be argument can be further simplified, you are most welcome to provide me.

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The following more general fact is true: if $G$ is a compact Lie group and $V$ is a faithful representation of it, then any finite-dimensional representation of $G$ occurs as a direct summand of $V^{\otimes n} \otimes V^{\ast \otimes m}$ for some $n, m$. For a proof see this MO question.

For application to this question take $G = SU(\ell+1)$, which has the same finite-dimensional representation theory (over $\mathbb{C}$) as $\mathfrak{sl}_{\ell+1}$. The standard representation $V$ is faithful, and moreover it has the property that $V^{\ast}$ occurs as a direct summand of a tensor power of $V$. In fact we have that $\wedge^k V$ is a direct summand of $V^{\otimes k}$, and

$$\wedge^{\ell} V \cong V^{\ast}$$

(because $\wedge^{\ell+1} V$ is trivial).