 Solution

If $\alpha \ne \beta$, the operator $f-\beta I$ is invertible on $W_{\alpha ,n}$. Indeed, there it equalswhose inverse is the finite geometric series

Now use induction on $d$. If ${\sum }_{i=1}^d{v}_i=0$, apply $(f-{\alpha }_{d}I{)}^n$. The $v_d$ term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in $W_{{\alpha }_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so $v_1,\dots ,v_d$ are linearly independent.