We can write $$e^{-\Lambda(n)}=\prod_{d\mid n}d^{\mu(d)},$$ where $\mu(n)$ is the Möbius function and thus $\Lambda(n)$ is von Mangoldt's function. Then taking the product from $1$ to $N$ we've for the second Chebyshev's function $\psi(x)$ that $$e^{-N\psi(N)}= \prod_{n=1}^{N} \prod_{d\mid n} d^{\mu(d)},$$ and the Prime Number Theorem states that $$\lim_{N\to\infty} \left( \prod_{n=1}^{N} \prod_{d\mid n} d^{\mu(d)} \right)^{\frac{1}{N^2}} =\frac{1}{e}.$$
I don't know if previous exercise was in the literature, and when I was looking a comparison with different partial product I wrote the following.
Combining Abel's identity $$\prod_{k=1}^Nk^{\mu(k)}=\frac{N^{M(N)}}{e^{\int_{1}^N\frac{M(t)}{t}dt}},$$ where $M(n)$ is the Mertens function, and the Prime Number Theorem one has $$\lim_{N\to\infty} \left( \prod_{k=1}^Nk^{\mu(k)} \right)^{\frac{1}{N}}=\lim_{N\to\infty}e^{-\frac{1}{N}\int_{1}^N\frac{M(t)}{t}dt} $$
Question. Can you justify that the limit exists and is $1$, I ask $$\lim_{N\to\infty} \left( \prod_{k=1}^Nk^{\mu(k)} \right)^{\frac{1}{N}}=1?$$ Thanks in advance.
I don't know if it is right. I know that the order of magnitude of the Mertens function is an unsolved problem, but are known good bounds. In the other I don't know if my questions is, perhaps an easy problem of analysis, on asumption that this limit exists. Truly I don't understand nothing of this post from MathOverflow, I say that I don't know after read it (my english is bad and there is a discussion) if the limit $\prod_{k=1}^{\infty}k^{\mu(k)}$ there exists (?), and take the cited value (also I don't know if such unsolved problem involving the Mertens function, is an additional problem when I read the partial product of this last infinite product as the ratio that I wrote, if you know/want clarify to me these doubts you can add comments about this).