Why is $\lim\limits_{n\to\infty} \prod\limits_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$? Mathematica returns these somewhat striking (to me, at any rate) infinite product identities:
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$$
and
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{(k+1)^2}{n^3}\right) = \sqrt[3]{e}$$
With larger exponents, checking with software seems to take forever.
I'm making a bit of a leap here, but these results suggest the following closed form for positive integer $p$:
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = e^{\frac1{p+1}}$$


*

*How does one compute either or both of the first two limits?

*Does the conjecture hold?


In the general case, we have
$$\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = \frac{n^{p+1}+(n+1)^p}{n^{p+1}+1} \prod_{k=1}^n \left(1 + \frac{k^p}{n^{p+1}}\right)$$
and the coefficient of the product converges to $1$. From here, I rewrite the product as a sum of logarithms and attempt to rearrange terms to reveal a Riemann sum, but I have had no luck so far. For instance, in the case of $p=1$,
$$\begin{align*}
\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac k{n^2}\right) &= \exp\left(\lim_{n\to\infty}\sum_{k=1}^n \ln\left(1 + \frac k{n^2}\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty}\sum_{k=1}^n \left(\ln\left(n+\frac kn\right) - \ln(n)\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty}\left(\sum_{k=1}^n \ln\left(n+\frac kn\right) - n\ln(n)\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty} n \left(\frac1n \sum_{k=1}^n \ln\left(n+\frac kn\right) - \ln(n)\right)\right)\\[1ex]
\end{align*}$$
 A: Consider any real $p > 0$. Note that for any $x>0$, $$ x \ge \ln(1+x) \ge x - x^2.$$ Indeed, at $x=0$ we have equality, and looking at the derivatives of $x,\ln(1+x),x-x^2$ respectivelly, we get $1,\frac{1}{1+x},1-2x$ respectivelly, and since $1 \ge \frac{1}{1+x} \ge 1-2x$ for any $x>0$ the result follows.  Hence $$ \ln\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = \sum_{k=1}^n \ln\left(1 + \frac{(k+1)^p}{n^{p+1}}\right) $$ can be bounded from both sides by $$ \frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p} - \frac{1}{n^2} \sum_{k=1}^n \frac{(k+1)^{2p}}{n^{2p}}\le \ln\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \le \frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p}. $$
Let us firstly deal with $"$linear$"$ factor, i.e $$\frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p} = \frac{1}{n} \sum_{k=1}^n \left( \frac{k+1}{n}\right)^p \to \int_0^1 x^pdx = \frac{1}{1+p},$$ where we used the fact that $x \mapsto x^p$ is continuous, to ensure the latter converges to integral via Riemann's  Sum Theorem (formally, we need to write $$ \sum_{k=1}^n \left( \frac{k+1}{n}\right)^p = \sum_{k=1}^n \left( \frac{k}{n}\right)^p - \frac{1}{n^p} + \left(\frac{n+1}{n}\right)^p$$ and note that both the second and the last term multiplied by $\frac{1}{n}$ converge to zero. Now, we need to deal with quadratic factor, but we do it really the same way, that is $$ \frac{1}{n^2} \sum_{k=1}^n \frac{(k+1)^{2p}}{n^{2p}} = \frac{1}{n} \cdot \frac{1}{n}\sum_{k=1}^n \left( \frac{k+1}{n}\right)^{2p} \to \frac{1}{\infty} \cdot \int_0^1 x^{2p}dx = 0 \cdot \frac{1}{2p+1} = 0. $$
Hence $$ \ln \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right)$$ is bounded from above by sequence converging to $\frac{1}{1+p}$ and from below by sequence convering to $\frac{1}{1+p}-0 = \frac{1}{1+p}$. Hence by squeeze theorem, $$ \ln \prod_{k=1}^n \left( 1+ \frac{(k+1)^p}{n^{p+1}}\right) \to \frac{1}{1+p}.$$
Since function $x \mapsto e^x$ is continuous, we can take the limit outside and get $$ \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \to \exp\left( \frac{1}{1+p}\right).$$
