<s>[INST] Find the remainder of $51!$ when divided by $61$? Find the remainder of $51!$ when divided by $61$ ?

My Try :
By Wilson's theorem $60! ≡ −1\pmod{61}$
Then, I can write 
$(60)(59)(58)(57)(56)(55)(54)(53)(52)51!≡−1\pmod{61}$
$(−1)(−2)(−3)(−4)(−5)(−6)(−7)(−8)(−9)51!≡−1\pmod{61}$
$(362880)51!\equiv1\pmod{61}$

How can I proceed after this OR Is there any other approach ?
 [/INST] Continued from your working without evaluating $9!$ explicitly:
$$51!9!\equiv 1 \mod 61$$
$$51! 2(3)(4)(5)(6)(7)(8)(9) \equiv 1 \mod 61$$
Since $2(5)(6)=60$,
$$51! (60)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (-1)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (3)(4)(7)(8)(9) \equiv -1 \mod 61$$
Since $9(7)=63$,
$$51! (3)(4)(8)(63) \equiv -1 \mod 61$$
$$51! (3)(4)(8)(2) \equiv -1 \mod 61$$
Since $2(4)(8)=64$,
$$51! (3)(64) \equiv -1 \mod 61$$
$$51! (3)(3) \equiv -1 \mod 61$$
$$51! (9) \equiv -1 \mod 61$$
Let's do Euclidean algorithm to compute $9^{-1} \mod 61$:
$$61=9(6)+7$$
$$9=7+2$$
$$7=3(2)+1$$
Hence $$1=7-3(2)=7-3(9-7)=4(7)-3(9)=4(61-9(6))-3(9)=4(61)-27(9)$$
$$9^{-1}\equiv -27 \mod 61$$.
Hence $$51!\equiv 27 \mod 61$$
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