Tensor product of representation of $\mathrm{SU}(2)$ Identity

I have troubles to understand an equation, which was stated in a lecture. Consider the spin-j representation $$V_{j}$$ of $$\mathrm{SU}(2)$$ with its standard basis

$$\{\vert j,m\rangle\}_{-j\leq m\leq j}$$

Now, consider the fundamental representation $$V_{\frac{1}{2}}$$.It was then stated that

$$\vert\uparrow\rangle^{\otimes 2j}=\vert j,j\rangle$$ where $$\vert\uparrow\rangle=\vert 1/2,1/2\rangle$$ is one of the two basis vectors of $$V_{1/2}$$. I cannot see why this is the case. I know that $$V_{1/2}^{\otimes 2j}$$ is a submodule of $$V_{j}$$, as a consequence of the Clebsch-Gordan decomposition, but I do not see how to obtain the equality from above. I guess, the result has something to do with the Clebsch-Gordan coefficients.

You have it backwards: $$V_j$$ (dimension $$2j+1$$) is a submodule of $$V^{\otimes 2j}_{1/2}$$ (dimension $$2^{2j}$$).
The state $$|\uparrow, \ldots, \uparrow\rangle$$ is the highest weight state of both the $$V_j$$ module and the $$V^{\otimes 2j}_{1/2}$$. In other words it is the unique state with the highest $$m$$ spin value $$m=j$$ in both the larger and the smaller space. Applying step-down ladder opertors keeps the resulting states in $$V_j$$ so $$|\uparrow, \ldots, \uparrow\rangle$$ must coincide with $$|j,j\rangle$$.