Infinite sum of integrals less than $e^{||f||_1}$ So basically I have this function
$$g(x_1)=1+\int_{0}^{x_1}f(x_2)dx_2+\int_{0}^{x_1}f(x_2)\int_{0}^{x_2}f(x_3)dx_3dx_2+ \\ \int_{0}^{x_1}f(x_2)\int_{0}^{x_2}f(x_3)\int_{0}^{x_3}f(x_4)dx_4dx_3dx_2+\ldots$$ 
And I want to show that $|g(x_1)|\leq e^{\int_{0}^{\infty}|f(y)|dy}$, where $f$ is continuous, bounded, absolutely Riemann-integrable, etc. 
I get the general idea;  I want to eventually compare it to the Taylor series of $e^x$,... but it feels like nothing is working out! For instance if I let $M=\int_{0}^{\infty}|f(y)|dy$, then I could let each of the innermost integrals be $\leq M$ and keep going, but then I'd get $$1+M+M^2+M^3+\ldots$$ instead of the series I want. Or if I define $m=\sup{|f|}$, then I get $1+mx+m^2x^2/2+m^3x^3/3!+\ldots$ which seems closer, but it still isn't it. Is there something silly I'm missing?
 A: Just use the Grownwall's inequality. Then the result is trivial by definition. 
Edit : Perhabs you want more detail. Take a derivative of $g$, and by fundamental theorem of Calculus you get  $$g'(x_1)=f(x_1)+f(x_1)\int_0^{x_1}f(x_3)dx_3+f(x_1)\int_0^{x_1}f(x_3)\int_0^{x_3}f(x_4)dx_4dx_3+\cdots$$$$=f(x_1)\bigg(1+\int_0^{x_1}f(x_3)dx_3+\int_0^{x_1}f(x_3)\int_0^{x_3}f(x_4)dx_4dx_3+\cdots\bigg)$$$$=f(x_1)\bigg(1+\int_0^{x_1}f(x_2)dx_2+\int_0^{x_1}f(x_2)\int_0^{x_2}f(x_3)dx_2dx_3+\cdots\bigg)$$$$=f(x_1)g(x_1).$$ 
Therefore $$|g|'\leq|g'|=|f||g|,$$ we can apply Gronwall's inequality : $$|g(x)|\leq|g(0)|e^{\int_0^x|f(x)|dx}=e^{\int_0^x|f(x)|dx}.$$
A: By symmetry, 
\begin{align}
\int_{0}^{x_1}f(x_2)\int_{0}^{x_2}f(x_3)\int_{0}^{x_3}f(x_4)\,\mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2 &= \int_{0}^{x_1}\int_{0}^{x_1}\int_{0}^{x_1}f(x_2)f(x_3)f(x_4) 1_{x_4<x_3<x_2}\, \mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2 \\
&= \frac{1}{6} \int_{0}^{x_1}\int_{0}^{x_1}\int_{0}^{x_1}f(x_2)f(x_3)f(x_4) \, \mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2 \\
&= \frac{1}{6} \left( \int_0^{x_1} f(y) \, \mathrm{d}y \right)^3.
\end{align}
Here, the notation $1_{x_4<x_3<x_2}$ means : $\begin{cases} 1 &\text{if } x_4<x_3<x_2 \\ 0 &\text{else}\end{cases}$.
Edit : since @user780610 needs a bit more detail, here it goes. The cube $C=[0,x_1]^3$ can be separated into six pieces : namely,
\begin{align}
C_1 &= \{ (x_2,x_3,x_4) : x_2<x_3<x_4 \} \\
C_2 &= \{ (x_2,x_3,x_4) : x_2<x_4<x_3 \} \\
C_3 &= \{ (x_2,x_3,x_4) : x_3<x_2<x_4 \} \\
C_4 &= \{ (x_2,x_3,x_4) : x_3<x_4<x_2 \} \\
C_5 &= \{ (x_2,x_3,x_4) : x_4<x_2<x_3 \} \\
C_6 &= \{ (x_2,x_3,x_4) : x_4<x_3<x_2 \} \\
\end{align}
Here I am omitting bits that dont count towards integrals, for instance when $x_2=x_3$. Now our triple integral $I_6 := \int_{0}^{x_1}f(x_2)\int_{0}^{x_2}f(x_3)\int_{0}^{x_3}f(x_4)\,\mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2$ is precisely equal to 
$$\int_{C_6} f(x_2) f(x_3) f(x_4) \, \mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2.$$
Define the same way the integrals $I_1$, $\dots$ , $I_5$. By renaming the variables (that's what I meant by "symmetry") we easily see that $I_1 = \dots = I_6$. Adding them all up, we get
\begin{align*}
6 I_6 &= I_1 + \dots + I_6 \\
&= \int_{C} f(x_2) f(x_3) f(x_4) \, \mathrm{d}x_4\mathrm{d}x_3\mathrm{d}x_2 \\
&= \left( \int_0^{x_1} f(y) \, \mathrm{d}y \right)^3,
\end{align*}
which is what we were after.
More generally, the $n$-th term in the sum is equal to $\frac{1}{n!} \left( \int_0^{x_1} f(y) \, \mathrm{d}y \right)^n.$ Hence we have actually
$$g(x_1) = \exp\left(\int_0^{x_1} f(y) \,\mathrm{d}y \right).
$$
