Integral equation proof by induction I am trying to solve the following problem:

We define the following integral for $n \in \mathbb{N}$ and $x \in \mathbb{R}_+^*$:
$$ \mathcal{I}(n,x) = \mathcal{I}_x(n) = \int_0^Tt^ne^{-xt}dt $$
Prove that:
$$ \begin{align}
\mathcal{I}_x(n) & = \frac{n!}{x^{n+1}} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}}
\\[12pt]
& = \frac{n!}{x^{n+1}} \left(1 - e^{-xT}\sum_{k=0}^n{\frac{(xT)^k}{k!}}\right)\end{align} $$

My try through induction:


*

*Case $n=0$:
$$ \begin{align}
\mathcal{I}_x(0) & = \int_0^Te^{-xt}dt 
\\[12pt]
& = \left[-\frac{e^{-xt}}{x} \right]_{t=0}^{t=T} 
\\[12pt]
& = \frac{1}{x} - e^{-xT}\frac{1}{x}
\end{align} $$
Hence the statement is true for $n=0$.

*Assume case $n$ is true. Integration by parts gives us:
$$ \begin{align}
\mathcal{I}_x(n+1) & = \int_0^Tt^{n+1}e^{-xt}dt 
\\[12pt]
& = \left[-\frac{t^{n+1}e^{-xt}}{x} \right]_{t=0}^{t=T} + \frac{n+1}{x}\int_0^T{t^ne^{-xt}dt}
\\[12pt]
& = -\frac{T^{n+1}e^{-xT}}{x} + \frac{n+1}{x}\mathcal{I}_x(n)
\\[12pt]
& = -\frac{T^{n+1}e^{-xT}}{x} + \frac{n+1}{x}\left( \frac{n!}{x^{n+1}} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}} \right)
\\[12pt]
& = \frac{(n+1)!}{x^{(n+1)+1}}-\frac{T^{n+1}e^{-xT}}{x} - \frac{n+1}{x}e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{n-k+1}\frac{n!}{k!}} 
\\[12pt]
& = \frac{(n+1)!}{x^{(n+1)+1}}-\frac{T^{n+1}e^{-xT}}{x} - e^{-xT}\sum_{k=0}^n{T^k\left(\frac{1}{x}\right)^{(n+1)-k+1}\frac{(n+1)!}{k!}} 
\\[12pt]
& = \frac{(n+1)!}{x^{(n+1)+1}}- e^{-xT}\sum_{k=0}^{n+1}{T^k\left(\frac{1}{x}\right)^{(n+1)-k+1}\frac{(n+1)!}{k!}} 
\end{align} $$
Hence if the statement holds for $n$ it also holds for $n+1$, which completes the proof.
I was expecting a simpler expression rather than the complicated sum. Does my reasoning seem right here? Is there any mistake? Is there a way to simplify further the final expression?
 A: After looking around, it seems my result is accurate: for $n \in \mathbb{N}^*$ and $T \in \mathbb{R}_+^*$, let $\gamma(n,T)$ and $\Gamma(n,T)$ be the lower and upper incomplete gamma functions respectively:
$$ \begin{align}
& \gamma(n,T) = \int_0^Tu^{n-1}e^{-u}du
\\[12pt]
& \Gamma(n,T) = \int_T^{\infty}u^{n-1}e^{-u}du
\end{align} $$
Wikipedia's page on incomplete gamma functions states that $-$ if $n \in \mathbb{N}^*$:
$$ \Gamma(n,T) = (n-1)! \ e^{-T}\sum_{k=0}^{n-1} \frac{T^k}{k!} $$
Given $\Gamma(n) = \gamma(n,T) + \Gamma(n,T)$ and $\Gamma(n) = (n-1)!$ $-$ provided $n$ is a positive integer $-$ it comes:
$$ \gamma(n,T) = \Gamma(n) - \Gamma(n,T) = (n-1)! \left(1- e^{-T}\sum_{k=0}^{n-1} \frac{T^k}{k!}\right)$$
Hence, making the dependence on $T$ explicit by noting my integral $\mathcal{I}$ as $\mathcal{I}(n,x,T)$:
$$ \gamma(n,T) = \mathcal{I}(n-1,1,T) $$
A simple change of variable $-$ $u = xt$ $-$ in $\mathcal{I}(n,x,T)$ yields:
$$ \mathcal{I}(n,x,T) = \frac{1}{x^{n+1}}\int_0^{xT} u^n e^{-u}du = \frac{\gamma(n+1,xT)}{x^{n+1}}$$
Plugging this back into the sum representation of $\gamma(\cdot,\cdot)$:
$$ \mathcal{I}(n,x,T) = \frac{\gamma(n+1,xT)}{x^{n+1}} = \frac{n!}{x^{n+1}} \left(1- e^{-xT}\sum_{k=0}^n \frac{(xT)^k}{k!}\right)$$
which corresponds to my initial representation of $\mathcal{I}_x(n)$ so I guess it must be right $-$ given I now have 2 proofs.
