Show that
$\phi$ is the golden ratio
$H_n$ is the n-th harmonic number
$1,1,2,3,5...$ for $n=1,2,3,4,5,...$ is the n-th Fibonacci number
$$(\phi-1)^1(2-\phi)^{1\over4}(2\phi-3)^{1\over9}(5-3\phi)^{1\over16}\cdots(\phi F_{n-1}-F_n)^{1\over (n-1)^2}(F_{n+1}-\phi F_n)^{1\over n^2}=\phi^{-H_n}$$
I try:
I noticed that $$(-1)^{n+1}\phi F_n+(-1)^nF_{n+1}=\phi^{-n}$$
$$\phi^{-1}\phi^{-2\over4}\phi^{-3\over 9}\cdots\phi^{-n\over n^2}=\phi^{-(1+{1\over2}+{1\over3}+\cdots+{1\over n})}=\phi^{-H_n}$$
I believe this is correct.
Is there another way of proving this product?
Here's another way to prove your result.
Generalizing the product of the OP we define
$$p(n,a)=\prod_{k=1}^n \left((-1)^{k+1}a\; q(k, a) +(-1)^kq(k+1)\right)^{\frac{1}{k^2}}$$
with a real parameter $a$ and a set of functions $q(k, a)$, $k = 1, 2, 3, ...$
Assuming now that
$$q(k,a)=\frac{-(1-a)^n+a^n}{ 2a -1}$$
the general factor of the product becomes
$$(-1)^{k+1}a\; q(k, a) +(-1)^k\; q(k+1)$$
$$= (-1)^{k+1}a \frac{-(1-a)^k+a^k}{ 2a -1}+(-1)^k\frac{-(1-a)^{1+k}+a^{1+k}}{ 2a -1}$$
$$=\frac{(-1)^k}{ 2a -1}\left(- a \left(-(1-a)^k+a^k\right)-(1-a)^{1+k}+a^{1+k}\right)$$
$$=\frac{(-1)^k}{ 2a -1}\left(a(1-a)^k-a^{k+1}-(1-a)^{1+k}+a^{1+k}\right)$$
$$=\frac{(-1)^k}{ 2a -1}(a-(1-a)) (1-a)^k$$
$$=(-1)^k (1-a)^k$$
and, since $k$ is an integer, finally
$$=(a-1)^k$$
Hence the product becomes
$$p1(n,a) = \prod_{k=1}^n (a-1)^{k/k^2}= (a-1)^{\sum_{k=1}^n 1/k} = (a-1)^{H(n)} $$
and we have shown that the identity
$$p(n,a) = p1(n,a)$$
holds for any $a$ and integer $\text{n$>$=1}$.
The special case of OP is found by letting $\text{a -$>$ $\phi $}$ and $\text{q(n) -$>$ F(n)}$, and observing that $\text{$\phi $-1 = 1/$\phi $}$.
$Discussion$
The assumption of the form of the function $q$ was of course guided by the well-known explicit expression for the Fibonacci numbers (Binet formula) but it is more general.
Assuming $a>=1$ avoids certain difficulty with the definition of $(a-1)^{H(n)}$
In the limiting case $a\text{-$>$}1/2$ we have $q = 2^{1-n} n$