Sum to sum

We know that the sum of first $n$ natural numbers is: $\large\displaystyle\sum_{n=1}^{n}n=\dfrac{n(n+1)}{2}$ And, sum of cubes of first $n$ natural numbers is: $\large\displaystyle\sum_{n=1}^{n}n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

Therfore, $\large\left(\displaystyle\sum_{k=1}^{1001}k\right)^n=\left(\dfrac{k(k+1)}2\right)^n\quad\text{ and}\quad\left(\displaystyle\sum_{k=1}^{1001}k^3\right)^2=\left(\dfrac{k(k+1)}{2}\right)^4$

Equating both of these; $\large\left(\dfrac{k(k+1)}2\right)^n=\left(\dfrac{k(k+1)}{2}\right)^4\\ \implies \boxed{n=4}$