The problem you've shared involves proving a double sum identity involving binomial coefficients and powers of $x$ and $y$. The sum is expressed as:

$\sum_{i=1}^{n} \sum_{j=i}^{n} \binom{j}{i} \binom{n}{j} (x-1)^{j-i} (y-x)^{n-j} = y^n - (y-1)^n.$

This identity needs to be proven for $x, y \in \mathbb{R}$ and $n \in \mathbb{N}$.

Strategy to Approach the Problem

1. Simplify the Double Sum: We first analyze the double summation by simplifying the binomial terms and the powers of $(x-1)$ and $(y-x)$.

   Break down each part of the sum and see if it can be restructured into a more familiar form.

2. Try Expansion or Induction: This sum closely resembles binomial expansions, so using binomial expansion formulas for $(y-1)^n$ and $y^n$ might help.

3. Look for a Pattern: Since the right-hand side involves $y^n - (y-1)^n$, it might suggest that the left-hand side sums up terms from the binomial expansions of these terms.

4. Evaluate Base Cases: Check for small values of $n$, like $n = 1$ or $n = 2$, to see if the equation holds in simpler forms.

Would you like me to go through these steps in detail and work through the full proof?