Proving $ e^2 = e$?

Question

Given the function, $$ f(x)= \frac{1}{e^x-1} \prod_{n=1}^{\infty} \frac{x^n}{n!} $$

Then applying the log,

$$ \ln(f(x)) = \frac{1}{e^x-1} \sum_{n=1}^{\infty} \frac{x^n}{n!} = \frac{e^x-1}{e^x-1} = 1 $$

$$ \rightarrow \quad f(x) = e \quad \rightarrow \quad \prod_{n=1}^{\infty}\frac{x^n}{n!} =e(e^x-1) $$

$$ $$

For $x=1$ we get,

$$ \prod_{n=1}^{\infty} \frac{1}{n!} = \frac{1}{1!} \cdot \frac{1}{2!} \cdot \frac{1}{3!} \cdot ...\frac{1}{n!} \rightarrow 0$$

Which means, $$ e(e-1)=0 \quad \rightarrow \quad e^2=e$$

What have I done wrong?

Answer 1

You cannot take the natural log of $0$.

Answer 2

You take the logarithm of the function $f(x)$ which is everywhere equal to $0$ except for $x=0$ where it is undefined. Also, note that for $a= \frac{1}{e^x-1}$ and $b_n = \frac{x^n}{n!}$

$$ \log \left(a \prod b_n\right) = \log(a) + \sum \log(b_n) \not= a \cdot \sum b_n. $$