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?
 A: If you insist on doing it by induction, look at
$$p(k+1)-p(k)=5(k+2)(k+3)(k+4)(k+5).$$
Among the four consecutive numbers $k+2,k+3,k+4,k+5$, one number is a multiple of four, and another (different) number is a multiple of two. There is also at least one multiple of three. So $5(k+2)(k+3)(k+4)(k+5)$ is divisible by $5\times4\times2\times3=120$, which implies the result.
With very little adjustment, you can rephrase this argument without induction, which is probably the cleaner way to do it.
A: Let $D$ the operator on $\mathbb{Z}[x]$ given by $P\to D(P)(x)=P(x+1)$. It is easy to see that ($I$=identity) $D(P)-P=(D-I)(P)$ is of degree strictly less than $P$. Hence if $P$ is of degree $s$, $(D-I)^{s+1}=0$ ; in your case, $P$ is of degree $5$, hence $(D-I)^6(P)=0$, and you can see that $P(n+6)$ is a combination with constant coefficients in $\mathbb{Z}$ of $P(n),...P(n+5)$. Now your induction is easy (if for $m, m+1,m+2,m+3,m+4,m+5$ you have the good property, show it works for $m+1,...,m+6$) (note that you can begin with $-5,-4,-3,-2,-1, 0$)
A: $120=2\cdot 3\cdot 4\cdot 5$.  $\Rightarrow   120|1\cdot 2\cdot 3\cdot 4\cdot 5$ 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.
