Finding the Rank of a special $n\times n$ matrix

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Let $A=[a_{ij} ],1≤i ,j≤n$ such that $a_{ij}=max\{i,j\}$. Then the rank of matrix A is equal to ____

Here is what I tried
By putting $n=1,2,3...$ I came to know that A is invertible and hence It has rank $n$.
So, I took an $n\times n$ matrix and tried to convert It into Row echelon form. $$A=\begin{bmatrix} 1 & 2 & 3 \hspace{5 pt}... \hspace{5 pt} n \\ 2 & 2 & 3 \hspace{5 pt} ... \hspace{5 pt}n\\ ... & ... & ... \hspace{5 pt} \\ n & n &\hspace{8 pt} ...\hspace{5 pt}n \\ \end{bmatrix}$$ That did not work out, also It is very tedious to prove that the determinant is non-zero for all n.
Then I tried PMI but I couldn't prove that determinant is non-zero for $n=k+1$ if It is non-zero for $n=k$