Can anyone help me proving this with Mathematical induction?

Prove that the product of $n$ consecutive numbers is divisible by $n!$

E.g. for $n=5$, we have that $5! | (10\cdot11\cdot12\cdot13\cdot14)$

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Hint:

It is obvious that $n!\mid n!$ for any $n$.

It is also obvious that $1\mid k$ for any $k$.

Notice also that $(k+1)(k+2)\cdots(k+n)=k(k+1)\cdots(k+n-1) + n\cdot \underbrace{(k+1)(k+2)\cdots(k+n-1)}_{(n-1)~\text{consecutive numbers}}$

To prove: $n!$ divides the product of n consecutive integers. Proof: Consider n consecutive integers: $k$, $k+1$, $k+2$…, $k+n-1$. Since $1$ divides $k$, and $2$ divides either $k$ or $k+1$, and $3$ divides either $k$ or $k+1$ or $k+2$…, and $n$ divides either $k$ or $k+1$ or $k+2$… or $k+n-1$, therefore $1$ x $2$ x $3$… x $n$, = $n!$, divides k(k+1)(k+2)…(k+n-1).