The Gamma function: does $0\cdot\infty=0$? My course notes (Mathematics BSc, second year Statistics module, unpublished) have

For $n=1,2,\dots$, we have $\Gamma(n)=(n-1)!$


Proof. Note that $$\Gamma(1)=\int_0^\infty e^{-u}=[-e^{-u}]_0^\infty=1.$$


Using integration by parts, for any $\alpha>1$,
\begin{align*}
\int_0^\infty u^{\alpha-1}e^{-u} \text{d}u
& = \left[u^{\alpha-1}(-e^{-u})\right]_0^\infty - \int_0^\infty(\alpha-1)u^{\alpha-2}(-e^{-u}) \text{d}u \\
& = 0 + (\alpha-1)\int_0^\infty u^{\alpha-2}e^{-u} \text{d}u
\end{align*}


That is, $\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha-1)$. By repeatedly applying this formula, for any $n\in\mathbb{N}$ we have $\Gamma(n)=(n-1)(n-2)\dots(3)(2)\Gamma(1)$, and since $\Gamma(1)=1$ this gives $\Gamma(n)=n!$.

My question is: how come $\left[u^{\alpha-1}(-e^{-u})\right]_0^\infty=0$? I get
\begin{align*}
\left[u^{\alpha-1}(-e^{-u})\right]_0^\infty
& = \left[\infty^{\alpha-1}(-e^{-\infty})\right] - \left[0^{\alpha-1}(-e^{-0})\right] \\
& = \infty \cdot 0 - 0 \cdot 0 \\
& = \text{undefined} - 0 \\
& = \text{undefined}.
\end{align*}
What have I missed?
 A: Since this question was migrated, I will try to answer it in a fundamental way. OP asks why
$$\left[u^{\alpha-1}(-e^{-u})\right]_0^\infty=0$$
Hiding in this question though is an even more basic question: what do we mean by
$$\int_{0}^\infty f(x)dx$$
If you know anything about Riemann integration (which I believe you may not just based on your question) then this is a non-sensical thing to ask as Riemann integrals are defined over intervals; that is
$$\int_a^bf(x)dx$$
To extend this to limits of integration at infinity, we define
$$\int_{0}^\infty f(x)dx:=\lim_{a\to\infty}\int_0^a f(x)dx$$
Then your original question becomes: what is
$$\lim_{x\to\infty}\left[u^{\alpha-1}(-e^{-u})\right]_0^x$$
This we can do:
$$\left[u^{\alpha-1}(-e^{-u})\right]_0^x=x^{\alpha-1}(-e^{-x})-0^{\alpha-1}(-e^{-0})=-x^{\alpha-1}e^{-x}$$
This implies
$$\lim_{x\to\infty}\left[u^{\alpha-1}(-e^{-u})\right]_0^x=-\lim_{x\to\infty}x^{\alpha-1}e^{-x}$$
Let $N$ be the smallest integer greater than $\alpha$. Then for $x>0$ we have
$$x^{1-\alpha}e^x=x^{1-\alpha}\sum_{n=0}^\infty \frac{x^n}{n!}>x^{1-\alpha}\frac{x^N}{N!}=\frac{x^{N+1-\alpha}}{N!}$$
By definition, we have $N+1-\alpha>\alpha+1-\alpha=1$. Thus, the expression is greater than $\frac{x}{N!}$. This implies
$$\lim_{x\to \infty}x^{1-\alpha}e^x\geq \lim_{x\to \infty}\frac{x}{N!}=\infty$$
$$\Rightarrow \lim_{x\to\infty}x^{\alpha-1}e^{-x}=0$$
