Step 1: Check whether A itself is singular.
Look at the rows of $A$: row1 $-2\times$row2 $+$ row3 gives the zero row, because the entries in each row are just consecutive integers shifted by one. So the rows of $A$ are linearly dependent, meaning $|A|=0$.
Step 2: Use this to handle the whole expression at once.
Since $\left|A^{2026}-A^{2025}\right|=\left|A^{2025}(A-I)\right|=|A|^{2025}\,|A-I|$, and we already know $|A|=0$, the factor $|A|^{2025}$ is zero.
Step 3: Conclude without even touching A - I.
Multiplying anything by zero gives zero, so the determinant of the whole expression is zero regardless of what $|A-I|$ works out to be.
\[ \boxed{0} \]