Is $f^{(0)}:=f$ a valid assumption for a proof involving derivatives on induction?

My motivation for this was proving the general Cauchy-Integral formula (in complex analysis) for an arbitrary derivative. Every book I read shows at least the first derivative using a $\delta-\epsilon$ argument, but we already did this technique/style when showing the proof for $f(z_0)=\int_\Gamma \frac{f(z)}{(z-z_0)}dz$.

So my question is a general one (i.e. I mention the C-Int formula only as my motivation), and I'm curious if (in the general case), we may (when convenient) consider the "zeroth derivative" (the function) as a base case for induction on a derivative.... obviously, the general formula for any such problem wasn't probably "seen" by looking at only that base case (I'm certain that the historical derivation for the general formula required the first three derivatives since the third derivative is the first time $n!≠n:n≠0$. None the less, I'm wondering if "zero derivative" is a valid assumption for such a proof or just a "definition".

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That is the definition of $f^{(0)}$. It is a reasonable definition, since it is consistent with the formula $(f^{(m)})^{(n)}=f^{(m+n)}$.

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Right... I agree, but its almost as if doing such a thing in a proof that relies on evaluating derivatives, will then never actually involve computing a derivative, because the base case isn't "really" a derivative and then with the inductive step you assume the result and to close the proof then you never even compute a derivative (using a limit argument) in order to show something about derivatives. Its kinda ironic, right? – Squirtle Sep 6 '12 at 1:43
To prove something about the $0$-th derivative, you will have to know or prove something about the function. – André Nicolas Sep 6 '12 at 1:45
Thank you.... that makes sense.... especially in light of the fact that I said I used the delta-epsilon argument in evaluating the Cauchy Integral formula (for zero derivative). – Squirtle Sep 6 '12 at 1:54