During watching a video course about Laplace-transform, I faced the following integral, $$\int_{0}^{\infty} t^6 \, e^{-2t} \, dt$$ The teacher compared it with the $L\left\{f(t)\right\} = \int_{0}^{\infty} f(t) \, e^{-st} \, dt$ and said here $f(t) = t^6$ and $s=2$. Hence, $L\left\{t^6\right\} = \dfrac{6!}{s^7}$ and plugging in $s=2$, gives $\dfrac{6!}{2^7}$.
But when I evaluate the integral with usual method I get a different answer,
$$\int_{0}^{\infty} x^6 \, e^{-2t} \, dt = x^6 \, \int_{0}^{\infty} e^{-2t} \, dt = \left. \dfrac{x^6 \, e^{-2t}}{-2} \right|_{0}^{\infty} = \dfrac{x^6}{2}. $$ I'm wondering why I got a different value? Is there something wrong with this approach?
Both integrals follow from: \begin{align} I_{n}(s) &= \int_{0}^{\infty} e^{- s t} \, t^{n} \, dt \\ &= \int_{0}^{\infty} e^{-u} \, \left(\frac{u}{s}\right)^{n} \, \frac{du}{s} \hspace{5mm} \text{where} \quad u = s t \\ &= \frac{1}{s^{n+1}} \, \int_{0}^{\infty} e^{-u} \, u^{(n+1)-1} \, du \\ &= \frac{\Gamma(n+1)}{s^{n+1}}. \end{align} This is the same as saying $$ L\left\{ t^{n} \right\} = \frac{n!}{s^{n+1}} $$ with the standard definition of the Laplace transform.
In the example of $n = 6$ and $s = 2$ the result is $$ \int_{0}^{\infty} e^{-2 t} \, t^{6} \, dt = \frac{6!}{2^{7}}. $$