Why is $(n+1)(n+2)\cdots(n+s)$ divisible by $s!$? Why is
$$(n+1)(n+2) \cdots (n+s), \text{ with } n,s \in ℕ$$
divisible by $s!$ ?
I derived this from the formula that gives the number of combinations of class $k$ with $n$ elements:
$$\frac{n!}{k!(n-k)!}$$
 A: Every sequence of $m$ consecutive integers has one member divisible by $m$. For example, two consecutive integers contain one integer divisible by $2$, three consecutive integers contain one integer divisible by $3$, and so on. $s$ consecutive integers not only has one integer divisible by $s$, but it also contains at least one sub sequence of $s-a$ integers for all $1\le a <s$. Thus a sequence of $s$ consecutive integers, regardless of the starting point $n$, will have integers in it divisible by every integer $1\le a\le s$, so the product of the members of that sequence will be divisible by $s!$
A: Let $N = (n+1)(n+2)\cdots (n+s)$. We must show that for each prime $p$, the number of times $p$ occurs in the prime factorization of $N$ is at least that the number of times $p$ occurs in $s!$.
Let $m_i$ be the number of entries in the list $L=(n+1,n+2,\dots,n+s)$ which are divisible by $p^i$. You can convince yourself that the number of factors of $p$ in $N$ is 
$$
m_1 + m_2 + m_3 + \dots
$$
I claim that 
$$
m_i \ge \lfloor s/p^i\rfloor
$$
This follows because, breaking $L$ into $\lfloor s/p^i\rfloor$ disjoint sections of $p^i$ consecutive integers (with some numbers left over), each section contains exactly one multiple of $p^i$. Therefore, the number of $p$'s in $N$ is at least
$$
\lfloor s/p\rfloor + \lfloor s/p^2\rfloor + \lfloor s/p^3\rfloor + \dots
$$
But you can also show that the above is exactly the number of occurences of $p$ in $s!$, by the same reasoning. This completes the proof.
A: We know that the number ways to pick $s$ items from $n + s$ choices must be $\frac {(n + 1)(n+2).....(n+s)}{s!}=\frac {(n+s)!}{n!}$ by basic combinatorics so $s!$ must divide $(n + 1)(n+2).....(n+s)$.
It used to astonish me that a practical explanation was much simpler than an algebraic proof.  It galled my actually.  
An algebraic proof should be simple:  As there are $s$ consecutive terms $s$ will divide exactly one of the terms.  Then $s-1$ will divide one of the terms (at least) and $s-2$ will divide one of the terms.  And so on so $s!$ should divide the product.
The trouble that as $k$ will divide a term and $m$ will divide a term, it is possible that they divide the same term and we could be over counting.
Consider a prime $p\le  s$ and suppose ... well will do this step by step:
If $p \le s < 2p$ then $p^1|s!$ is the highest power that divide $s!$ and of the $s\ge p$ consecutive terms, one is divisible by $p$ so $p|\frac {(n+s)!}{n!}$.  And the highest fact of $p$ that divides $s!$ divide $\frac {(n+s)!}{n!}$ and we can move on to the next prime factor of $s!$.  We don't have to worry about double counting because once we are done with $p$ we will be dealing with terms relatively prime to $p$.
If $k*p \le s < (k+1)*p \le p^2$, then there are $k$ terms in $1,... s$ that are divisible by $p$ and $p^k|s!$ is the highest term that divide $s!$.  Now $p$ must divide one of the terms $n+1,...., n+p$ terms.  Call that term $m$ and $p$ must divide each of the terms $m+p....., m+ (k-1)p \le n+ kp \le n+s$.  So $p^k|\frac {(n+s)!}{n!}$.
And if $k*p^m + j*p^{m-1} + .... \le s$.  We can repeat the above steps to get that there are $k$ terms in $n+1,... ,n+s$ that are $v, v+p^m,... , v + (k-1)p^m$ that are divisible by $p^m$.  of the remaining terms there are $j$ terms that are each divisible by $p^{m-1}$ and so on.  Thus $p^{km + j(m-1) + ...}|s!$ is the highest power of $p$ that divides $s!$ and $p^{km + j(m-1) + ...}|\frac {(n+s)!}{n!}$.
And that's it.  We won't double count by divide all prime factors of $s!$ out of $\frac {(n+s)!}{n!}$.
