The hermitian conjugate has the folowing properties $$ a_{\alpha} = ( a_{\alpha}^{\dagger} )^{\dagger}. $$ Finally we found $$ a_\alpha\underbrace{|\alpha_1'\alpha_2' \dots \alpha_{n+1}'}\rangle_{\neq \alpha} = 0, \quad \textrm{in particular } a_\alpha |0\rangle = 0, $$ and $$ a_\alpha |\alpha\alpha_1\alpha_2 \dots \alpha_{n}\rangle = |\alpha_1\alpha_2 \dots \alpha_{n}\rangle, $$ and the corresponding commutator algebra $$ \{a_{\alpha}^{\dagger},a_{\beta}^{\dagger}\} = \{a_{\alpha},a_{\beta}\} = 0 \hspace{0.5cm} \{a_\alpha^{\dagger},a_\beta \} = \delta_{\alpha\beta}. $$