How to prove that $(a+b)!/a!b!$ is an integer

Question

My approach was show that with,

$a!= \prod_{i=1}^{X} p_i^{a_i}$ , $b! = \prod_{i=1}^{Y} p_i^{b_i}$ and $(a+b)!= \prod_{i=1}^{Z} p_i^{c_i}$ WLOG suppose $b \geq a$ and clearly $a+b \geq a,b$ thererfore $Z \leq Y \leq X$ (note because they are factorials they share the same prime factors for example for $i\leq X$ the $p_i's$ are the same for $a$, $b$ and $a+b$, $a's$ set of prime factors is a subset of $b's$ and $b's$ set of prime factors is a subset of $(a+b)!'s$

1. The set of prime factors for $a!b!$ is a subset of the set of prime factors of $(a+b)!$

2. For $a!b! = \prod_{i=1}^{Y} p_i^{f_i}$ I need to show that $f_i \leq c_i$ for all $i$

For 1:

$a!b! = \prod_{i=1}^{X} p_i^{a_i} \cdot \prod_{i=1}^{Y} p_i^{b_i}$ $= (p_1^{a_1} \cdot p_2^{a_2} \cdot . . . \cdot p_X^{a_X})$ $(p_1^{a_1} \cdot . . . p_x^{b_x} \cdot p_{x+1}^{b_{x+1}} \cdot . . . p_Y^{b_Y})$ $= (p_1^{a_1+b_1} \cdot p_2^{a_2+b_2} \cdot . . . p_X^{a_X+b_X} \cdot p_{x+1}^{b_{x+1}} \cdot . . . p_Y^{b_Y}$) since $Y \leq Z$ we know that the set of prime factors of $a!b!$ is a subset of $(a+b)!$ set of prime factors.

For 2: We know that $a_i = \lfloor \frac{a}{p_i} \rfloor$ $+ \lfloor \frac{a}{p_i^{2}} \rfloor + . . . + \lfloor \frac{a}{p_i^{r}} \rfloor$ $r$ is defined by $p^r \leq a < p^{r+1}$

$b_i = \lfloor \frac{b}{p_i} \rfloor$ $+ \lfloor \frac{b}{p_i^2} \rfloor + . . . + \lfloor \frac{b}{p_i^{l}} \rfloor$ $l$ is defined by $p^l \leq b < p^{l+1}$

$c_i = \lfloor \frac{a+b}{p_i} \rfloor$ $+ \lfloor \frac{a+b}{p_i^{2}} \rfloor + . . . + \lfloor \frac{a+b}{p_i^{k}} \rfloor$ $k$ is defined by $p^k \leq a+b < p^{k+1}$

This is were I get stuck I want to show that $f_i \le c_i$ for all i by using the fact that $\lfloor \alpha \rfloor + \lfloor \beta \rfloor \leq \lfloor \alpha + \beta \rfloor$.

However I don't know how many terms $a_i, b_i$ or $c_i$ My guess on how roughylgo from here is as follows: I guess $b$ probably has more terms than $a$ and $a+b$ very likely has more terms than both of them added together (i.e $k \geq l \geq r$), so equipped with these assumptions I try to show that $a_i+b_i \leq \lfloor \frac{a+b}{p_i} \rfloor$ $+ \lfloor \frac{a+b}{p_i^{2}} \rfloor + . . . + \lfloor \frac{a+b}{p_i^{r}} \rfloor$ $+ \lfloor \frac{b}{p^{r+1}} \rfloor + . . . + \lfloor \frac{b}{p_i^{l}} \rfloor$ $\leq \lfloor \frac{a+b}{p_i} \rfloor$ $+ \lfloor \frac{a+b}{p_i^{2}} \rfloor + . . . + \lfloor \frac{a+b}{p_i^{k}} \rfloor$ .