Prove that $(a+1)(a+2)...(a+b)$ is divisible by $b!$ [duplicate]

You can do induction simultaneously on $a$ and $b$.

Basis. If $a=0$ the statement is clearly true. If $b=0$ it is also true (an empty product is equal to one).

Inductive hypothesis. Assume that the statement is true if we decrease $a$ or $b$ by one. More precisely, let $L(a,b)$ denote the statement for given $a$ and $b$. We assume that $L(a-1,b)$ and $L(a,b-1)$ are true.

Inductive step. Write: $$ \begin{align} & (a+1) \ldots (a+b-1) (a+b) \\ = \, a & (a+1) \ldots (a+b-1) \\ + \quad & (a+1) \ldots (a+b-1) b \\ \end{align} $$ The first term divides $b!$ because $L(a-1,b)$ is true. The second term divides $b!$ because $L(a,b-1)$ is true.

$$ \frac{(a+b)(a+b-1)(a+b-2)\cdots(a+1)}{b(b-1)(b-2)\cdots3\cdot2\cdot1}=\binom{a+b}{b} $$ Since binomial coefficients are listed in Pascal's Triangle, they are integers.

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$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$