Formula for Repeated Integration of Exponentials I am struggling with proving the following equation: 
$$
k!\int_0^t \int_0^{t-s_1}\cdot\cdot\cdot\int_0^{t-\sum_{i=1}^{k-1} s_i} \prod_{j=1}^{k}e^{-js_j} ds_k \cdot\cdot\cdot ds_1 
=(1-e^{-t})^{k}
$$
for arbitrary $k\in\mathbb{N}$. I tried to use induction and evaluated the most inner integral, which generates a 
$$
e^{-k\left(t-\sum_{i=1}^{k-1} s_i\right)}.
$$
More precisely, I assume that the formula is true for $n< k$, then 
$$
k!\int_0^t \int_0^{t-s_1}\cdot\cdot\cdot\int_0^{t-\sum_{i=1}^{k-1} s_i} \prod_{j=1}^{k}e^{-js_j} ds_k \cdot\cdot\cdot ds_1 
 \\
=(1-e^{-t})^{k-1}- e^{-t}(k-1)! \int_0^t \int_0^{t-s_1}\cdot\cdot\cdot\int_0^{t-\sum_{i=1}^{k-2} s_i} e^{-(k-1)t} \prod_{j=1}^{k-1}e^{-(k-j)s_j} ds_{k-1} \cdot\cdot\cdot ds_1 
$$
i.e. the second integral needs to evaluate to $(1-e^{-t})^{k-1}$, which would then finish the proof. 
Is there an easy way to evaluate this or even an easier way to prove the formula in general? Does there exist a generalised Cauchy formula for repeated integrals (to me it looks similar). Any help is appreciated, thanks a lot!
 A: Let $f_k(t) = k ! \int _ { 0 } ^ { t } \int _ { 0 } ^ { t - s _ { 1 } } \cdots \int _ { 0 } ^ { t - \sum _ { i = 1 } ^ { k - 1 } s _ { i } } \prod _ { j = 1 } ^ { k } e ^ { - j s _ { j } } d s _ { k } \cdots d s _ { 1 }$. Consider $f_k'(t)$:
$$
f_k'(t) = k! \left.\int _ { 0 } ^ { t - s _ { 1 } } \cdots \int _ { 0 } ^ { t - \sum _ { i = 1 } ^ { k - 1 } s _ { i } } \prod _ { j = 1 } ^ { k } e ^ { - j s _ { j } } d s _ { k } \cdots d s _ { 2 }\right|_{s_1=t} + \\
k! \int _ { 0 } ^ { t } \left.\int _ { 0 } ^ { t - s _ { 1 } -s_2} \cdots \int _ { 0 } ^ { t - \sum _ { i = 1 } ^ { k - 1 } s _ { i } } \prod _ { j = 1 } ^ { k } e ^ { - j s _ { j } } d s _ { k } \cdots d s _ { 3 }\right|_{s_2=t-s_1}ds_1
+ \ldots+\\
k! \left. \int_0^t\int _ { 0 } ^ { t - s _ { 1 } } \cdots \int _ { 0 } ^ { t - \sum _ { i = 1 } ^ { k - 2 } s _ { i } } \prod _ { j = 1 } ^ { k } e ^ { - j s _ { j } }\right|_{s_k=t-s_1\ldots-s_{k-1}}
 d s _ { k-1 } \cdots d s _ { 2 }
$$
All terms here except for the last one are zero since in one of the integrals both limits are $0$. The term inside the last integral is:
$$
\left.\prod _ { j = 1 } ^ { k } e ^ { - j s _ { j } }\right|_{s_k=t-s_1...-s_{k-1}} =
e^{-k(t-s_1\ldots-s_{k-1})}\prod _ { j = 1 } ^ { k-1 } e ^ { - j s _ { j } }
$$
The coefficient before the product:
$$
e^{-k(t-s_1\ldots-s_{k-1})}=1-k\int_0^{t-s_1\ldots-s_{k-1}}e^{-ks_k}ds_k.
$$
Collecting everything together, we get:
$$
f_k'(t)=kf_{k-1}(t) - kf_k(t).
$$
Now we can use induction. We directly check that $f_1(t)=1-e^{-t}$. Then we assume that $f_{k-1}(t)=(1-e^{-t})^{k-1}$. Then we get differential equation, that satisfies Cauchy conditions:
$$
f_k'(t) = k(1-e^{-t})^{k-1} + kf_k(t)
$$
with initial condition $f_k(0) = 0$. We show with direct calculation that function $y(t)=(1-e^{-t})^k$ satisfies the differntial equation and the initial condition. Finally, Cauchy–Kowalevski theorem states that this equation has the unique solution, which we have found.
Edit A simpler approach.
One can notice that integration is done over a simplex on unit vectors, which is symmetric under any permutation of variables. So it's possible to change the order of integration:
$$
\int_0^t\int_0^{t-s_1}\dots\int_0^{t-s_1\ldots-s_{k-1}} F ds_k\ldots ds_1 = \\
\int_0^t\int_0^{t-s_k}\int_0^{t-s_k-s_1}\dots\int_0^{t-s_k-s_1\ldots-s_{k-1}} F ds_{k-1}\ldots ds_1ds_k. 
$$
Thus, we can simply write:
$$
f_k(t) = k\int_0^t e^{-ks_k}f_{k-1}(t-s_k) ds_k = k\int_0^t (1-e^{-t+s_k})^{k-1}e^{-ks_k}ds_k = \\
k \int_0^t (e^{-s_k}-e^{-t})^{k-1} e^{-s_k}ds_k = 
-k\int_1^{e^{-t}} (u-e^{-t})^{k-1}du = -\left.(u-e^{-t})^k\right|_{1}^{e^{-t}}=(1-e^{-t})^k
$$
