Wilson's Theorem will help

By a corollary of Wilson's theorem (2) , if $p$ is a prime of the form $4k+1$ , where $k$ is an integer, then $[(2k)!]^2 \equiv -1 \pmod p$ . Since $101$ is of the form $4k+1$ , where $k=25$ , then $(50!)^2 \equiv -1 \equiv \boxed{100} \pmod {101}$ .

Since $101$ is prime, by Wilson's Theorem , $100!\equiv -1\pmod{101}$ We observe that $100!=100 \times 99\times 98\times \cdots \times 51\times (50)!$ $100\equiv-1\pmod{101}$ $99\equiv -2\pmod{101}$ $98\equiv -3\pmod{101}$ $\cdot$ $\cdot$ $51\equiv -50\pmod{101}$ $50!\equiv 50!\pmod{101}$ Multiplying the above congruence relations, $100!\equiv (50!)^2\equiv -1\pmod{101}$ Therefore, $\boxed{(50!)^2\equiv100\pmod{101}}$