Proof check: Expectation of squared sum of independent mean zero random vectors

Question

Let $Z_1,...,Z_k$ be independent mean zero random vectors in $R^n$. Show that \begin{equation*} E\left\| \sum_{j=1}^{k} Z_j \right\|_2^2 = \sum_{j=1}^{k} E \left\| Z_j \right\|_2^2 \end{equation*} Proof: Let $X=Z_1+...+Z_k$ and $X^i,Z_j^i$ the ith component of $X,Z_j$. \begin{equation} E\left\| \sum_{j=1}^{k} Z_j \right\|_2^2=E [(X^1)^2+...(X^k)^2]=\sum_{i=1}^k E[(\sum_{j=1}^k Z_j^i)^2] \end{equation} Then for a fixed $i$ \begin{equation} (\sum_{j=1}^k Z_j^i)^2=\sum_{j=1}^k (Z_j^i)^2+R. \end{equation} The $R$ will consist of summands $Z_x^i Z_y^i$ with $x\neq y$. Because our random vectors are independent (this means the single components are also independent, right ?) we have $E[R]=0$. After some simple calculations the statement follows. Is this correct ?