Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$. 
Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.

I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5)
\\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$
I got stuck over here.
Am I right till here? Can someone tell what to do next?

$120=2\cdot 3\cdot 4\cdot 5$.  $\Rightarrow   120|1\cdot 2\cdot 3\cdot 4\cdot 5$ and $120|2\cdot 3\cdot 4\cdot 5\cdot 6$.
Now, if $120|k(k+1)(k+2)(k+3)(k+4)$ and $120|(k+1)(k+2)(k+3)(k+4)(k+5)$, then,
$120|(k+1)(k+2)(k+3)(k+4)\{(k+5)-k\}$
$\Rightarrow 24|(k+1)(k+2)(k+3)(k+4)$
Now, use normal induction and proceed.
