Cauchy's formula for repeated integrals proof by induction

We always have$$\int_a^b\frac{\mathrm d\varphi}{\mathrm dy}\,\mathrm dy=\varphi(b)-\varphi(a),$$by the Fundamental Theorem of Calculus. So,$$\int_a^x\frac{\mathrm d}{\mathrm dy}\left(\int_a^y(y-t)^nf(t)\,\mathrm dt\right)\,\mathrm dy=\int_a^x(x-t)^nf(t)\,\mathrm dt-\overbrace{\int_a^a(a-t)^nf(t)\,\mathrm dt}^{\phantom{0}=0}.$$

I´m going to show a sketch of a proof for Cauchy´s formula for repeated integrals. The formula says: $$f^{(-n)}(x)=\frac{1}{(n-1)!}\int_{a}^{x} (x-t)^{n-1}f(t)dt$$ Let start with $n=3$, and later on we try to generalize it. Then: $$f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu $$ $$\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu=\int_{a}^{x}\left[ \int_{a}^{u} \int_{a}^{s}f(t)dtds\right]du $$ Now concentrate in the double integral inside the parenthesis. Here is a sketch of it´s region of integration.

We can swap the order the integration, according to the following picture:

If we do so, we get the following expression for the inner double integral $$ \int_{a}^{u} \int_{a}^{s}f(t)dtds=\int_{a}^{u} \int_{t}^{u} f(t)dsdt$$ $$\int_{a}^{u} \int_{t}^{u} f(t)dsdt=\int_{a}^{u}f(t)dt \int_{t}^{u}ds=\int_{a}^{u}(u-t)f(t)dt$$

$$ \Rightarrow f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu =\int_{a}^{x} \left[\int_{a}^{u}(u-t)f(t)dt\right]du$$

Proceeding in the same way we did before (changing the order of integration), we get $$\int_{a}^{x} \int_{a}^{u}(u-t)f(t)dtdu=\int_{a}^{x} \int_{t}^{x}(u-t)f(t)dudt$$ $$\int_{a}^{x} \int_{t}^{x}(u-t)f(t)dudt=\int_{a}^{x}f(t)dt \int_{t}^{x}(u-t)du$$ The second integral on the right hand side of the equation can be solved using a substitution: let $s=u-t \Rightarrow ds=du$ $$\int_{a}^{x}f(t)dt \int_{t}^{x}(u-t)du=\int_{a}^{x}f(t)dt\int_{0}^{x-t} sds$$ And finally we get: $$f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu =\int_{a}^{x} \frac{(x-t)^{2}}{2!}f(t)dt$$ It´s starting to emerge a pattern, if we keep integrating further, we may prove by induction Cauchy´s formula.