Let $1\leq n\in \mathbb{N}$. Let $\mathcal{E}:=(e_1, \ldots , e_n)$ the standard basis of $\mathbb{R}^n$. We define $\displaystyle{e_i':=\sum_{j=1}^ie_j}$.
I want to show also that $(e_1', \ldots , e_n')$ is a basis of $\mathbb{R}^n$.
For that we use induction.
Base case: Do we start from $n=3$ that we have above or from $n=1$ ?
Inductive Hypothesis: We assume that it holds for $n=k$.
Inductive Step: We want to show that it holds for $n=k+1$. Could you give me a hint how we can show that if we have an additional element and a higher dimension we get again a basis?
It suffice to prove $\mathrm{rank}\{e_1', \cdots , e_n'\}=n$.
Consider the linear transformation bewteen these two sets of vectors:$$(e_1', \cdots , e_n')=(e_1, \cdots , e_n)\begin{pmatrix} 1 & 1 & 1 & \cdots & 1\\ 0 & 1 & 1 & \cdots & 1\\ 0 & 0 & 1 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 0 & 0& 0 & \cdots & 1 \end{pmatrix}=(e_1, \cdots , e_n)A.$$ Note that $\det A=1 \neq 0$. Therefore $\mathrm{rank}\{e_1', \cdots , e_n'\}=\mathrm{rank}\{e_1, \cdots , e_n\}=n$.