I just wanted to write the iterated version of:
$$ a^2 - b^2 = (a+b)(a-b)$$
As:
$$ n^x - 1 = (n^{ \sqrt{x}} +1) \color{red}{(n^{ \sqrt{x}} -1)} = \\ (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1) \color{red}{(n^{ \sqrt{\sqrt{x}}} -1)} = \\ (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) \color{red}{(n^{ \sqrt{\sqrt{\sqrt{x}}}} - 1)} = \dots $$
And so:
$$n^x - 1 = (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) ... \Rightarrow \\n^x - 1 = \prod_{k=1}^{\infty} (n^{x^{\left (\frac{1}{2k}\right)}} + 1)$$
But when plugging $n=1$ we get:
$$ 1^x - 1 \equiv 0 = \prod_{k=1}^{\infty} (1^{x^{\left (\frac{1}{2k}\right)}} + 1) \equiv \prod_{k=1}^{\infty} (1+1) = \prod_{k=1}^{\infty} 2 = \infty$$
Where did I go wrong? I think that I "omitted" the $n^{x^{(\frac{1}{2k})}} - 1$ part, but isn't it valid when we talk about infinite series? I would appreciate if you could clear things up.
Thank you!
Should be instead $$n^x-1 = (n^{x/2}-1)(n^{x/2}+1)$$, as in general,
$$n^x-1 \not = (n^{\sqrt{x}}-1)(n^{\sqrt{x}}+1).$$
This was your error.