prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$  prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$
Let $P(n)$ be the proposition we want to prov, ie: $P(n):=24 \mid(n)(n+1)(n+2)(n+3)$
For $P(1)$ we have: $24 \mid(1)(1+1)(1+2)(1+3)\implies6 \mid(1)(2)(3)(4)\implies24 \mid24$, so $P(1)$is true.
For $P(2)$ we have: $24 \mid(2)(2+1)(2+2)(2+3)\implies6 \mid(2)(3)(4)(5)\implies24 \mid120$, so $P(1)$is true
Inductive Hypothesis: Let $n=k$ and we assume that $P(k):=24\mid k(k+1)(k+2)(k+3)$ is true.
Inductive Step:
$$(k+1)(k+2)(k+3)(k+4)$$
$$k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$
Using the assumption of $P(k) \implies \exists a\in \mathbb Z$, such that, $(k+1)(k+2)(k+3)=24\cdot a$
so: $$=24\cdot a +4(k+1)(k+2)(k+3)$$

Continuing from where you left, we just need to prove that for $n=k+1$, $P(n)$ is an integer multiple of $24$.
$$P(k+1) = (k+1)(k+2)(k+3)(k+4)$$
$$P(k+1) = k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$
$1st$ term on right hand side is $P(k)$ which is an integer multiple of $24$ from your inductive hypothesis.
$2nd$ term on the right hand side has a product of $3$ consecutive integers and hence divisible by $6$. So on a whole divisible by $4*6=24$.
On a whole the right hand side is divisible by $24$. Hence $P(k+1)$ is an integral multiple of $24$.
