How to show $\{u_1,\sum_i^2{u_i},\cdots,\sum_i^n{u_i}\}$ is linearly independent if $\{u_1,{u_2},\cdots,{u_n}\}$?

Suppose $(\lambda_1,...,\lambda_n)$ is a set of coefficients such that $$ 0 = \sum_{i=1}^n \lambda_iv_i \tag{1} $$ where $v_i = \sum_{k=1}^i u_k$. We want to show that then one must have $ \lambda_i=0$ for all $1\leq i\leq n$.

We can rewrite this as $$ 0 = \sum_{i=1}^n \lambda_i \sum_{k=1}^i u_k= \sum_{k=1}^n \sum_{i=k}^n \lambda_i u_k = \sum_{k=1}^n \left(\sum_{i=k}^n \lambda_i\right) u_k \tag{2} $$ and so, by independence of the $u_k$'s, we must have $$ \sum_{i=k}^n \lambda_i = 0 \qquad \forall 1\leq k\leq n \tag{3} $$ Show that this implies $ \lambda_i=0$ for all $1\leq i\leq n$ (e.g., by induction, taking $k$ from $n$ to $1$ in (2)).

---

To prove the converse: assume, again letting $v_i = \sum_{k=i}^n u_i$, that $(v_1,\dots,v_n)$ is linearly independent. Let $\alpha_1,\dots,\alpha_n$ such that $$ 0 = \sum_{i=1}^n \alpha_i u_i $$ Noting that $u_i = v_i - v_{i+1}$ (for $1\leq i<n$, we get $$ 0 = \alpha_n v_n + \sum_{i=1}^{n-1} \alpha_i v_i - \sum_{i=1}^{n-1} \alpha_i v_{i+1} = \alpha_n v_n + \sum_{i=1}^{n-1} \alpha_i v_i - \sum_{i=2}^{n} \alpha_{i-1} v_{i} = \alpha_1 v_1 + \sum_{i=2}^{n} (\alpha_i - \alpha_{i-1}) v_{i} $$ By our assumption of linear independence of the $v_i$'s, we have $0 = \alpha_1 = \alpha_i - \alpha_{i-1}$ (for all $i$). Again by induction, this implies that $\alpha_i = 0$ for all $i$. That shows that $(u_1,\dots,u_n)$ is linearly independent.