Can this be solved for $f(n)$? While working upon a partial sum formula for the harmonics, I came across a necessity for the function defined below
$f\left(\frac{n(n-1)}{2} +1\right) = n!$
Can it be solved for $f(n)$?

Let your formula define $f(t)$ for all integer $t$ on the form $t=\frac{n(n-1)}{2}+1$ with $n$ integer. For other $t$ define $f$ to be whatever you want. The resulting function will satisfy your equation. This means that there is no unique function satisfying your equation.
However, there is a 'natural' function. Solving $\frac{n(n-1)}{2}+1 = t$ for $t$ we get
$$n = \frac{1 \pm \sqrt{8t-7}}{2}$$
and by using $n! = \Gamma(n+1)$ for integer $n$ where $\Gamma$ is the gamma function we get (by taking the positive branch)
$$f(t) = \Gamma\left(\frac{1 + \sqrt{8t-7}}{2} + 1\right)$$
which is a real function for all $t\geq \frac{7}{8}$.
