Long chain of remainders $\pmod{11}$

but isn't the answer 4? I doubt it is true, by multiplying!

Actually, you can solve it even without using Wilson's theorem. Just evaluate the value of $10! \mod 11$ by splitting up $10!$ and finding the remainder of each part and then find the remainder at last of the value obtained if it is $>10$ like this,

$10!=10\times 9\times 8\times 7\ldots 2\times1 = 90\times 56\times 30\times 24 \\ \implies 10! (\mod 11) = (90\times 56\times 30\times 24) (\mod 11)\\ \implies 10! (\mod 11) \equiv (2\times 1\times 8\times 2) (\mod 11) \\ \implies 10! (\mod 11) \equiv 32 (\mod 11) \equiv 10 (\mod 11)\\ \implies \boxed{10! (\mod 11) \equiv 10 (\mod 11)}$