jmc

Let $q(x)=xp(x)-1.$ Then $q(x)$ has degree at most 9. Also, $p(n)=n\cdot p(n)-1=0$ for $n=1,$ 2, 3, $\dots ,$ 9, so $$q(x)=c(x-1)(x-2)\cdots (x-9)$$for some constant $c.$

We know that $q(0)=0\cdot p(0)-1=-1.$ Setting $x=0$ in the equation above, we get $$q(0)=-9!\cdot c,$$so $c=\frac{1}{9!}.$ Hence, $$q(x)=\frac{(x-1)(x-2)\cdots (x-9)}{9!}.$$Then $q(10)=\frac{9\cdot 8\cdots 1}{9!}=1,$ so $$p(10)=\frac{q(10)+1}{10}=\frac{2}{10}=\boxed{\frac{1}{5}}.$$