Reducing multiple integrals to single integral In our integral equations course, we derived the following equation
$$
\int_0^x\int_0^{x_1} ...\int_0^{x_{n-1}}u(x_n)dx_ndx_{n-1}...dx_1=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1}u(t)dt
$$
Then my professor said that it is easy to see from this, that
$$
\int_0^x\int_0^{x_1} ...\int_0^{x_{n-1}}(x-x_n)u(x_n)dx_ndx_{n-1}...dx_1=\frac{1}{n!}\int_0^x(x-t)^{n}u(t)dt
$$
This same result is given in our textbox
However, when I  checked the last equation with some simple example ($u(t)=1,n=2$) it seems that the factor $\frac{1}{n!}$ should be $\frac{1}{(n-1)!}$. Is this correct?
$$
\frac{1}{2!}\int_0^x (x-t)^2 \, dt=\frac{x^3}{6}
$$
$$
\int _0^x\int _0^{x_1}(x-x_2)dx_2dx_1=\frac{x^3}{3}
$$
$\frac{1}{2!}$ should be $\frac{1}{1!}$.
 A: You have
$$\int_0^x\int_0^{x_1} ...\int_0^{x_{n-1}}f(x_n)dx_ndx_{n-1}...dx_1=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1}f(t)dt \tag{1}$$
Integrate both sides from $0$ to $u$
$$\int_0^u\int_0^x\int_0^{x_1} ...\int_0^{x_{n-1}}f(x_n)dx_ndx_{n-1}...dx_1\,dx=\frac{1}{(n-1)!} \int_0^u\int_0^x(x-t)^{n-1}f(t)\, dt \, dx \tag{2}$$
Now, focus on the R.H.S. of $(2)$
$$\begin{aligned}
\frac{1}{(n-1)!} \int_0^u\int_0^x(x-t)^{n-1}f(t)\, dt \, dx &=\frac{1}{(n-1)!} \int_0^u\int_t^u(x-t)^{n-1}f(t)\, dx \, dt\\
 &=\frac{1}{(n-1)!} \int_0^u f(t)\int_t^u(x-t)^{n-1}\, dx \, dt\\
&=\frac{1}{(n-1)!} \int_0^u f(t)\int_0^{u-t} w^{n-1}\, dw \, dt\\
&=\frac{1}{(n-1)!} \int_0^u f(t)\left(\frac{w^n}{n}\Big|_0^{u-t} \right) \, dt\\
&=\frac{1}{n!} \int_0^u (u-t)^n f(t) \, dt\\
&=\frac{1}{n!} \int_0^x (x-t)^n f(t) \, dt \qquad \blacksquare\\
\end{aligned}$$
Then $(2)$ becomes
$$\int_0^x\int_0^{x_1}\int_0^{x_2} ...\int_0^{x_{n}}f(x_{n+1})dx_{n+1}dx_{n}...dx_1\,dx=\frac{1}{n!} \int_0^x (x-t)^n f(t) \, dt \tag{3}$$
Note that the L.H.S. of $(1)$ has $n$ integrals and that the the L.H.S. of $(2)$ has $n+1$ integrals
A: Yes, as stated in your question the second formula should have a factor of $\frac{1}{(n-1)!}$. To see this we will rewrite the first equation as
$$\int_0^x \int_0^{x_1}\cdots\int_0^{x_{n-1}}f(x_n)dx_n dx_{n-1}\cdots dx_1=\frac{1}{(n-1)!}\int_0^x(x-t)^{n-1}f(t)dt$$
and let $f(t)=(x-t)u(t)$ then we obtain
$$\int_0^x \int_0^{x_1}\cdots\int_0^{x_{n-1}}(x-x_n)u(x_n)dx_n dx_{n-1}\cdots dx_1=\frac{1}{(n-1)!}\int_0^x(x-t)^{n}u(t)dt.$$
However, as I wrote in a comment, this isn't really equation (1.130) from your textbook.
