A question regarding $\lim\limits_{n\to\infty} n(n+1)(n+2)\dots(n+r)$. We know that $\lim\limits_{n\to\infty} \frac{n(n+1)(n+2)\dots(n+r)}{n^{r+1}}$, where $r$ is a finite positive integer, is $1$. 

Hence, $\lim\limits_{n\to\infty} n(n+1)(n+2)\dots(n+r)=\lim\limits_{n\to\infty}{n^{r+1}}$.

This should imply that $\lim\limits_{n\to\infty} \left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=0$

However, $$\left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=n^r(1+2+\dots+n)+n^{r-1}(1.2+1.3+\dots+(r-1)r)+\dots+r!$$

Shouldn't $\lim\limits_{n\to\infty}n^r(1+2+\dots+n)+n^{r-1}(1.2+1.3+\dots+(r-1)r)+\dots+r!$ tend to infinity?

Thanks in advance!
This is false: "This should imply that $\lim\limits_{n\to\infty} \left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=0$"

We know $\lim_{n=1}^\infty \frac{n+1}{n}=1$, yet we know $n+1\neq n$.  All the limit tells you is that the terms with the highest power are the same, it doesn't tell you anything about the terms with lower powers.
Hint:

$$\ln\left(\frac{n(n+1)...(n+r)}{n^{r+1}}\right)=\sum_{i=0}^{r}\ln\left(1+\frac{i}{n}\right)\to 0$$

Hence $$\lim_{n\to \infty}\frac{n(n+1)...(n+r)}{n^{r+1}}=1 $$
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We have $$\prod_{i=0}^{n}(n+i)=n(n+1)...(n+r)=n^{r+1}+r!\left(\frac{1}{0!}+\frac{1}{1!}+...+\frac{1}{r!}\right)n^r+...+r!$$

$$\to \frac{n(n+1)...(n+r)}{n^{r+1}}=1+r!\left(\frac{1}{0!}+\frac{1}{1!}+...+\frac{1}{r!}\right)\frac{1}{n}+O\left(\frac{1}{n}\right)\to 1$$

Hence $$\lim_{n\to \infty}\frac{n(n+1)...(n+r)}{n^{r+1}}=1 $$