Convergence of quadratic variation of Brownian Motion

Question

Let $(\Omega,\mathcal{F},\mathbb{F}=\{\mathcal{F}_t:t>0\},P)$ be a filtred probability space and $B=\{B_t:t>0\}$ a Brownian Motion. Say $0=t_0^k<...<t_{N_k}^k=t$ is a partition of $[0,t]$ such that $\lim_{k\to\infty}\{\sup_{j=1,...,N_k}|t_j^k-t_{j-1}^k|\}=0$. I'm trying to prove that: $$\lim_{k\to\infty}\lVert{V_k^2(B)_t-t}\rVert_{L^2}^2=0,$$ where: $$V_k^2(B)_t=\sum_{j=1}^{N_k}(B_{t_{j}^k}-B_{t_{j-1}^k})^2,$$ and: $$\lVert{X}\rVert_{L^2}^2=E[|X|^2]=\int_{\Omega}|X|^2dP.$$ My professor wrote that $\forall{i<j}$ we have: $$E[(B_{t_j^k}-B_{t_{j-1}^k})^2(B_{t_i^k}-B_{t_{i-1}^k})^2]=\\=E\{E[(B_{t_j^k}-B_{t_{j-1}^k})^2|\mathcal{F}_{t_{j-1}^k}](B_{t_i^k}-B_{t_{i-1}^k})^2\}=\\=E[(B_{t_j^k}-B_{t_{j-1}^k})^2]\cdot{E[(B_{t_i^k}-B_{t_{i-1}^k})^2]},$$ but I don't get why these equations hold.