Example 7 Find the polynomial $P(x)$ that satisfies $x P(x-1)=(x-26) P(x)$.

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Analysis: $x=0,1, \cdots, 25$ are all roots of $P(x)$, so $P(x)=x(x-1) \cdots(x-25) Q(x)$, and $Q(x)$ $=Q(x-1)$. If the degree of $Q(x)$ is $n>0$, then it has $n$ complex roots $x_{1}, \cdots, x_{n}$, and $x_{1}+1, \cdots, x_{n}+1$ are also its $n$ roots, the sum of which are not equal, a contradiction! Therefore, $P(x)=a x(x-1) \cdots(x-25)$.