I have some problem to see how we can solve this problem.
Consider $e_1=( 1, 0, ..., 0)\in\mathbb{R}^d$, $e_2=( 0, 1, ..., 0)\in\mathbb{R}^d$, ..., $e_d=( 0, 0, ..., 1)\in\mathbb{R}^d$. We are interested in the matrix $A_k=\sum_{i=1}^{k} e_ie_i^{t}$
We would like to find his rank for all $k\in(1,..., d)$. My background in linear algebra is mainly in vector space / sub-space and linear mapping. My aptempt : The idea is to use the fact that a matrix can always be express in terms of linear mapping so I will look at the subspace spanned by this linear mapping. For $k=1$ we have that $ A_1 =(a_{ij})$ with $a_{i,j}=1$ for $i = j =1$ and zero otherwise. To find the rank of $A_1$ we need to find the dimension of {$v\in\mathbb{R}^{d} : v = (a, 0, ..., 0)$ for $a\in\mathbb{R}$} clearly this is 1.
Then for $A_2$ we have that $A_2=a_{i j}$ with $a_{i j} =$ for $i=j=1, 2$ and zero otherwise. To find the rank of $A_2$ we have to find the dimension of {$v\in\mathbb{R}^{d} : v = (a, b, ..., 0)$ for $(a,b)\in\mathbb{R}^2$} but $v=(a,b, ..., 0)=a(1, 0, ..., 0) + b(0, 1, ..., 0)$
so the dimension is 2 and therefore the rank of $A_2$ is 2.
Given this pattern, I suppose this is true for $k=n <d$ and I will try to show that it is true for $k=n+1$. We have that $A_{n+1}=a_{ij}$ where $a_{ij}=1$ for $i,j=1,2,...n+1$ and zero otherwise. Thus we want to find the dimension of {$v\in\mathbb{R}^{d} : v = (a_1, a_2, ..., a_{n+1},..., 0)$ for $(a_1, a_2, ..., a_{n+1})\in\mathbb{R}^{n+1}$} but $v=a_1(1, ..., 0)+a_2(0,1, ..., 0) + ... + a_{n+1}(0, ..., 1, ..., 0)$
and we know that the first $n$ vectors are linearly independents, clearly the $n+1$th vectors is linearly independent from the others. We conclude that it is a basis so the dimension of the subspace spanned by $A_{n+1}$ is $n+1$ and his rank is $n+1$. We conclude this is true for all $k\in(1,..., d)$. So $A_d$ is of rank d which makes sense since it is the identity matrix of size $d$.
Is this correct ? Is there another way to do it ?
Thank you a lot