A Problem Of Convergence

These kinds of limits can be reduced nicely with the root-to-ratio limit:

$$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$

In your case this becomes:

\begin{align}L&=\lim_{n\to\infty}\frac{m+n+1}{n+1}\prod_{k=1}^n\frac{m+k}{n+1}\frac n{m+k}\\&=\lim_{n\to\infty}\prod_{k=1}^n\frac n{n+1}\\&=\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\\&=\lim_{n\to\infty}\left[\left(1-\frac1{n+1}\right)^{n+1}\right]^{n/(n+1)}\\&=e^{-1}\end{align}

Your hint is the way to go here. Let's do it that way. Your hint can be stated as

Lemma 1: Let $\{a_n\} $ be a sequence of positive terms such that $a_n\to L$ as $n\to\infty $. Then $$\sqrt[n] {a_1a_2\dots a_n} \to L$$ as $n\to \infty $.

The key here is to choose $a_n$ properly. Let us choose $$a_n=\left. \prod_{i=1}^{n+1}\frac{m+i}{n+1}\middle/\prod_{i=1}^{n}\frac{m+i}{n}\right.$$ so that $$a_n=\frac{m+n+1}{n+1}\cdot\frac{n^n}{(n+1)^n}$$ so that $a_n\to 1/e$.

Next note that $$a_1a_2\dots a_n=\frac{1}{m+1}\prod_{i=1}^{n+1}\frac{m+i} {n+1}$$ and by Lemma 1 the $n$-th root of above expression tends to $1/e$. Since $\sqrt [n] {m+1}\to 1$ it follows that $$\left(\prod_{i=1}^{n+1}\frac{m+i}{n+1}\right) ^{1/n}\to \frac{1}{e}$$ Raising the expression on left to power $n/(n+1)$ we can see that the above equation holds if the exponent $1/n$ is replaced with $1/(n+1)$. And then we can replace $n$ by $n-1$ everywhere to see that the desired limit is $1/e$.

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You should be able to show that lemma 1 above is equivalent to the root-to-ratio limit result used in another answer here.