There is a unique log-convex $f:[0,\infty)\to(0,\infty)$ that satisfies $f(x+1) = (x+1)f(x)$ and $f(0)=1$ (namely $f(x) = x! = \Gamma(x+1)$).

Here's a generalization that I think is easier to prove:

If $\delta:[0,\infty)\to\mathbb{R}$ is such that $\delta^{(n)}$ decreases monotonically to zero, then there is a unique $f$ such that $f(x+1) = f(x) + \delta(x)$, $f(0) = 0$, and $f^{(n)}$ is monotonically increasing.

(To recover the original statement, take $n=1$ and $\delta(x) = \log(x+1)$.)

The benefit of this formulation is that it can be proved by induction on $n$. The inductive step is pretty routine and involves applying the $(n-1)$-case to $f'$ and $\delta'$. The basis case for $n=0$ still requires a nontrivial argument, but to me it feels much simpler and more intuitive than the case $n=1$ (for the original theorem), and it's actually easy to remember (in contrast to the direct proof of the Bohr-Mollerup theorem, which I find hard to remember). All you have to do is apply the functional equation $k$ times to get $f(x+k)-f(k) = f(x) + \sum_{j=0}^{k-1}[\delta(x+j)-\delta(j)]$ and take the limit as $k$ approaches infinity; you end up with an increasing function of $x$ that is zero whenever $x$ is an integer and is thus zero for all $x$. Thus $f(x) = \sum_{j=0}^{\infty}[\delta(j) - \delta(x+j)]$ is the unique solution. (Since $\delta$ is decreasing, we do have an increasing function, and the sum converges since $\delta(x)\to 0$ as $x\to +\infty$.)