# 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 $(f^{(m)})^{(n)}=f^{(m+n)}$.
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