Do you know any nontrivial analytic function f(z) with zeros only at positive integer values of the argument z = 1, 2, 3, 4, ... ? If yes, please give some example.
PS: I already thought of $f(x)=\frac{1}{\Gamma(-x+1)}$. Any other nice options?
EDIT: To avoid trivial solutions due to restriction of definition range, please consider the required function to be defined in the whole complex plane.
On
Denote $(p \in \mathbb{N}\cup \{0\})$ $$E\left({z},\, p\right)=\begin{cases}1-z, & p=0, \\ (1-z)e^{z+\frac{z^2}{2}+\ldots+\frac{z^p}{p}}, & p\geqslant{1}. \end{cases}$$ Function $E\left({z},\, p\right)$ is called by primary or elementary Weierstrass factor of genus $p$. Let $\{a_\nu \} \subset \mathbb{C}$ be a sequence of non-zero complex numbers such that $|{a_\nu}| \rightarrow \infty.$ Then function $$f(z)=\prod\limits_{\nu=1}^{\infty}{E\left(\dfrac{z}{a_\nu},\, p\right)}$$ called as Weierstrass canonical product, is entire function that has zeroes at points $a_\nu$ and only at them.
On
$$ f(x) = 3^x - 2^x $$ has 0 as a zero. $$ f(x) = 4^x - 2 \cdot 3^x + 2^x$$ has 0 and 1 as a zero. $$ f(x) = 5^x - 3\cdot4^x + 3\cdot3^x - 2^x $$ has 0, 1, and 2 as a zero. $$ f(x) = 6^x - 4\cdot5^x + 6\cdot4^x - 4\cdot3^x + 2^x $$ has 0, 1, 2, and 3 as a zero.
See the pattern. Continuing
$$ f(x) = 7^x - 5\cdot6^x+10\cdot5^x-10\cdot4^x+5\cdot3^x-2^x$$ has 0, 1, 2, 3, and 4 as a zero.
$$ f(x) = 8^x-6\cdot7^x+15\cdot6^x-20\cdot5^x+15\cdot4^x-6\cdot3^x+2^x $$ has 0, 1, 2, 3, 4, and 5 as a zero.
The pattern is the base decreases by one. Signs alternate. The coefficients are the numbers of the Pascal's triangle.
All other entire functions $f$ with simple zeros exactly at positive integers differ from your $f_o(z)=1/\Gamma(-z+1)$ by a function of the form $e^{g(z)}$ for entire $g$, and vice-versa. That is, $f(z)=f_o(z) \cdot e^{g(z)}$ for entire $g$, and vice-versa. Indeed: $f(z)/f_o(z)$ has no zeros and is entire, so is of the form $e^{g(z)}$, since we can define its logarithm.