If $I_n(a) = \int_0^a (e^x - 1)^n \, \mathrm{d}x$, show $I_n(a) = \frac{(e^a - 1)^n}{n} - I_{n-1}(a)$. Question:

If
$$I_n(a) = \int_0^a (e^x - 1)^n \, \mathrm{d}x,$$
where $a$ is a real constant, show that $I_0(a) = a$ and by writing $(e^x - 1)^n = (e^x - 1)(e^x - 1)^{n - 1}$ obtain the reduction formula
$$I_n(a) = \frac{(e^a - 1)^n}{n} - I_{n-1}(a).$$

I'm struggling to get to the answer. Have I made any mistakes so far?
\begin{align*}
\require{cancel}
I_n(a) &= \int_0^a (e^x - 1)(e^x - 1)^{n - 1} \, \mathrm{d}x \qquad \begin{aligned}v' &= e^x - 1 \\ v &= e^x - x\end{aligned} \quad \begin{aligned}u &= (e^x - 1)^{n-1} \\ u' &= (n-1)(e^x - 1)^{n-2}(e^x)\end{aligned} \\
&= \left[(e^x - 1)^{n-1}(e^x - x)\right]^a_0 - \int_0^a (n - 1)(e^x)(e^x - 1)^{n-2} \, \mathrm{d}x \\
&= \left[(e^a - 1)^{n-1}(e^a - a)\right] - \left[(e^0 - 1\cancel{)^{n-1}}\cdot e^0 - 0\right] - \int_0^a (n - 1)(e^x)(e^x - 1)^{n-2} \, \mathrm{d}x \\
&= \left[(e^a - 1)^{n-1}(e^a - a)\right] - \int_0^a (n - 1)(e^x)(e^x - 1)^{n-2} \, \mathrm{d}x \\
\end{align*}
 A: As far as I can see your working is correct, but you have gone down the wrong path. Think about your line:
$$I_n(a) = \int_{0}^{a}(e^x-1)(e^x-1)^{n-1}$$
What will happen if we apply IBP? Well as you've seen the expression, we get from this isn't very useful. Let's look at the result we want. It has a $-I_{n-1}(a)$ term so we want to look for a way to create this term in our expression. We can notice that a way to quickly get this term is to split up the integral like so:
$$I_n(a) = \int_{0}^{a}e^x(e^x-1)^{n-1} - \int_{0}^{a}(e^x-1)^{n-1} = \int_{0}^{a}e^x(e^x-1)^{n-1} - I_{n-1}(a)$$
Then:
$$\int_{0}^{a}e^x(e^x-1)^{n-1} =  \bigg [\frac{(e^x-1)^n}{n}\bigg]^{a}_0=\frac{(e^a-1)^n}{n}$$
So:
$$I_n(a)=\frac{(e^a-1)^n}{n} - I_{n-1}(a)$$
as required. 
Note: I've evaluated the integral above by recognition but if you find this confusing it would be easy to with the substitution $u=e^x-1$.
A: Consider the functions $f(a)=I_n(a)+I_{n-1}(a)$ and $g(a)=\frac{(e^a-1)^n}{n}$. Since $(e^a-1)^n$ is continuous on $\mathbb R$, the function $f$ is differentiable and $f'(a)=(e^a-1)^n+(e^a-1)^{n-1}=e^a(e^a-1)^{n-1}$.
On the other hand, $g$ is differentiable on $\mathbb R$ and by the Chain Rule, $g'(a)=(e^a-1)^{n-1}e^a=f'(a)$. Therefore, $f(a)=g(a)+C$ for every $a$ in $\mathbb R$.
But $f(0)=0$ and $g(0)=0$, so $C=0$. It follows that $f=g$, that is
$$
I_n(a)+I_{n-1}(a) = \frac{(e^a-1)^n}{n}
$$
