Proof that matrix has v1 in its nullspace, and v2 to vn are eigenvectors of Q2

Suppose we have a matrix V given by

The columns of V are orthogonal to each other. We have proved that $$V^TV = I$$. We also have proved that the columns of V form a basis for RN, and that $$$$ = $$a$$ 1 if $$b = a1v1+ ... +anVn$$.

Now we are given Q, and its eigenvalues such that

We've proved that the eigenvectors of Q are given by the columns of V.

Now we have to show that if $$Q_2 = Q - λ_1 v_1 v_1^T$$, $$v_1$$ is in Nul($$Q_2$$) and $$v_2 ... v_n$$ are eigenvectors

My approach so far was substituting $$Q_2=Q - λ_1 v_1 v_1^T$$ in $$Q_2 x =_? 0$$ But when I distribute $$v_1$$ through inner products, I'm not sure if I can say that $$$$ are 0. I mean, I know that $$$$ because they're orthonormal, but can I make the other claim? Do I have the right approach to this whole proof?

You are interested in the null space of $$Q_2$$ and the eigenvectors with non-zero eigenvalues. You are correct that you may obtain the null space by solving the system of equations $$Q_2 x = 0$$; however, this may be challenging in general. You actually already have a guess for the nullspace, so you should use this guess!
Here's a hint to get you started: to show that $$v_1$$ is in the null space of $$Q_2$$, try multiplying $$Q_2 v_1$$ and see what you get. Next, to show that $$v_2, \dots v_n$$ are eigenvectors of $$Q_2$$, try multiplying $$Q_2 v_i$$ and see what you get for $$i=2, 3, \dots n$$.
Bonus: If indeed $$v_i$$ is an eigenvector for $$Q_2$$ with $$i = 2,3, \dots n$$, then what is its eigenvalue?
Note that $$Q_2v_1 = (Q- \lambda_1 v_1 v_1^T)v_1 = Qv_1 - (\lambda_1 v_1v_1^T)v_1 = \lambda_1v_1 - \lambda_1v_1 (v_1^T v_1) = \lambda_1v_1 - \lambda_1 v_1 \cdot 1 = 0$$ where the second last equality follows from the fact that $$v_1^T \cdot v_1 = 1$$ (as the vectors are orthonormal, this follows from $$V^TV = I$$)
In the same way, you can prove that $$Q_2v_i = \lambda_iv_i$$ for each $$i \in \{2, \ldots, n\}$$ since $$v_1^Tv_i = 0$$ (orthonormal vectors).