<s>[INST] find function $f$ such that $f(x)=xf(x-1)$ and $f(1) = 1$ 
Find function $f$ such that  $f(x)=xf(x-1)$ and $f(1) = 1$.

I can prove that there is just one function as $f$ (see Proof1).
I know that there exists a pi function $\Pi(z) = \int_0^\infty e^{-t} t^z\, dt$ that fits as $f$ so it is the only solution.
My problem is, although I know the answer to be $\Pi(z)$ I can not reach it myself. I mean how can I find $f$ without knowing the answer before? how can I reach $\Pi(z)$ from the equation $f(x)=xf(x-1)$ and $f(1) = 1$?

Proof1
suppose $f(x)$ and $g(x)$ are two solutions to $f(x)=xf(x-1)$ and $f(1) = 1$
$$
\frac{f(x)}{g(x)} = h(x) \Rightarrow  \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g^2(x)} = h'(x) \\
\Rightarrow \frac{f(x-1) + xf'(x-1)}{g(x)} - \frac{f(x)}{g(x)}\frac{g(x-1) + xg'(x-1)}{g(x)} = h'(x) \\
\Rightarrow \frac{f(x-1)}{g(x)}+\frac{xf'(x-1)}{g(x)} - \frac{f(x)}{g(x)}(\frac{g(x-1)}{g(x)}+\frac{ xg'(x-1)}{g(x)}) = h'(x) \\
\Rightarrow \frac{1}{x}\frac{f(x)}{g(x)}+\frac{xf'(x-1)}{xg(x-1)} - \frac{f(x)}{g(x)}(\frac{1}{x}+\frac{xg'(x-1)}{xg(x-1)}) = h'(x) \\
\Rightarrow \left [\frac{1}{x}\frac{f(x)}{g(x)} - \frac{f(x)}{g(x)}\frac{1}{x}\right ] + \left [\frac{xf'(x-1)}{xg(x-1)} - \frac{f(x)}{g(x)}\frac{xg'(x-1)}{xg(x-1)}\right ] = h'(x) \\
\Rightarrow 0 + \left [\frac{\not{x}f'(x-1)}{\not{x}g(x-1)}\frac{g(x-1)}{g(x-1)} - \frac{\not{x}f(x-1)}{\not{x}g(x-1)}\frac{\not{x}g'(x-1)}{\not{x}g(x-1)}\right ] = h'(x) \\
\Rightarrow \frac{f'(x-1)g(x-1) - g'(x-1)f(x-1)}{g^2(x-1)} = h'(x) \\
\Rightarrow h'(x-1) = h'(x) \\
$$
now we know that $h(x)$ must be a line. But $h(x) = h(x+n)$ $^{h(x+1) = \frac{f(x+1)}{g(x+1)} = \frac{{x +1}f(x)}{{x +1}g(x)} = \frac{f(x)}{g(x)} = h(x)}$. So $h(x)$ must be a horizontal line so it is a constant. so:
$$
\frac{f(x)}{g(x)} = c, \frac{f(1)}{g(1)} = 1 \Rightarrow \frac{f(x)}{g(x)} = 1 \Rightarrow f(x) = g(x)
$$
 [/INST] No, it is not true.  Given any solution $f$ and any function $g$ that is periodic with period $1$ (i.e. $g(x+1) = g(x)$) and $g(1)=1$, then $h(x) = f(x) g(x)$ also satisfies your equations.  For example, you could take $g(x) = 1 + \sin(2\pi x)$.
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