Show that each number of $100!+1,100!+2,100!+3,...,100!+100$ is composite

$101$ is prime so by Wilson's theorem $100!\equiv -1\pmod {101}$ so $100!+1\equiv 0\pmod {101} $ so $101|100!+1$ .

For $1 <k\le 100$ we simply note the $k|100! $ (by definition) and $k|k $ so $k|100!+k $.

Note $N! +1$ might be prime for some $N $ but not if $N+1$ is prime. $3!+1=7$ is prime (but $3+1$ isnt).