I found the following simple but wrong derivation of the Gamma function:
We start from the definition of the exponential function $$ e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!} \\ \Rightarrow 1 = \sum_{k=0}^{\infty}\frac{x^k e^{-x}}{k!} $$ Now we want to find a continuous version of this expression and argue that since $k!$ is not defined for real numbers we have to replace it with some function $f(k)$ so that the following expression stays true $$ 1 = \int_{k=0}^{\infty}\frac{x^k e^{-x}}{f(k)} dk $$ So how do we solve for $f(k)$? We simply (and wrongly) exchange the variable of integration from $k \rightarrow x$ so we can move the unknown function f(k) to the left side and get $$ f(k) = \int_{x=0}^{\infty} x^k e^{-x} dx $$ This function is equal to $\Gamma[k+1]$. So we "calculated" very directly that the continuous version of the factorial is the $\Gamma$-function.
Of course this prove is wrong. But it seems so suggestive. So my question is, can one solve for $f(k)$ without any illegal tricks and derive the Gamma function in that way?