First of all, I'll repeat what I said in the comments: writing $\frac{x^h - 1}{h} = \frac{e^{h \log x} - 1}{h}$ makes it fairly clear that as $h \to 0$, this expression tends to $\log x$, and setting $h = a + 1$ this is precisely the desired result.

I should mention that this result is implicit in a certain fact well-known to people who do competition math, which is as follows. Given non-negative real numbers $x_1, ... x_n$, let $A_p(x_1, ... x_n) = \sqrt[p]{ \frac{x_1^p + ... + x_n^p}{n} }$ denote the $p$-power mean for $p \neq 0$. For $p = 0$, define $A_0(x_1, ... x_n) = \sqrt[n]{x_1 ... x_n}$ (the geometric mean, and also the limit as $p \to 0$ of the above).

**Theorem (Power Mean Inequality):** If $p \le q$, then $A_p \le A_q$.

If you like fancy keywords, then I will bring to your attention that as $p \to \infty$ the $p$-power mean approaches $\text{max}(x_1, ... x_n)$, which one can think of as the "low-temperature limit" of ordinary addition becoming tropical addition. Then $p \to 0$ can be thought of as the "high-temperature limit," in which ordinary addition becomes multiplication instead. Somebody who knows more statistical mechanics than I do (that is, any) can probably tell you the physical significance of this.