\(n=1\)のとき

\begin{align*} f^{\left(-1\right)}\left(x\right) & =\int_{a}^{x}f\left(y\right)dy\\ & =\left[\frac{1}{\left(n-1\right)!}\int_{a}^{x}\left(x-y\right)^{n-1}f\left(y\right)dy\right]_{n=1} \end{align*} となるので成り立つ。

\(n=k+1\)のとき、

\(n=k\)のとき成り立つと仮定する
\(n=k+1\)のとき
\begin{align*} f^{\left(-k+1\right)}\left(x\right) & =\int_{a}^{x}\int_{a}^{y_{1}}\cdots\int_{a}^{y_{k}}f\left(y_{k+1}\right)dy_{k+1}\cdots dy_{1}\\ & =\frac{1}{\left(k-1\right)!}\int_{a}^{x}\int_{a}^{y_{1}}\left(y_{1}-y_{2}\right)^{k-1}f\left(y_{2}\right)dy_{2}dy_{1}\\ & =\frac{1}{\left(k-1\right)!}\int_{a}^{x}\int_{y_{2}}^{x}\left(y_{1}-y_{2}\right)^{k-1}f\left(y_{2}\right)dy_{1}dy_{2}\\ & =\frac{1}{k!}\int_{a}^{x}\left(x-y_{2}\right)^{k}f\left(y_{2}\right)dy_{2} \end{align*} となるので\(n=k+1\)でも成り立つ。

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故に数学的帰納法より与式は成り立つ。