Finding limit of sequence: $\lim _{n \to \infty} {\frac{n!}{n^k(n-k)!}}=1$

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$k$ is nonnegative integer. I want to show that$$ \lim _{n \to \infty} {\frac{n!}{n^k(n-k)!}}=1$$

My try :

$$ \frac{n!}{n^k(n-k)!} = \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n}$$

I wanted use multiplicative rule

If $a_n \to a$ and $b_n \to b$ , then $a_n b_n \to ab$.

But It is impossible because $$ \frac{n!}{n^k(n-k)!} = \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n}$$ It is infinite product.

I want you to help me.