How would I get the laplace tranform of -5t^2 step by step

I have the identity that it equals n!/s^(n+1), but I don't understand how they got to it.... The factorial brings me to believe that I need the taylor series expansion of each, because it is a long and repetitive process to do ibp...

how would I do integral(from 0,infinity) of (t^n)e^(-st)dt step by step... Please and thanks

For s > 0 and n ≥ 0, let

I_n(s) = ∫[t=0 to ∞] t^n e^{-st} dt.

Let T > 0. By integration by parts,

∫[t = 0 to T] t^n e^{-st} dt

= t^n (-1/s e^{-st})[t = 0 to T] - ∫[t = 0 to T] (-1/s e^{-st}) (n t^{n-1}) dt

=-T^n e^{-sT}/s + n/s ∫[t = 0 to T] t^{n-1} e^{-st} dt.

Since s>0, T^n e^{-sT} → 0 as T → ∞. Therefore

∫[t = 0 to ∞] t^n e^{-st} dt = n/s ∫[t = 0 to ∞] t^{n-1} e^{-st} dt,

in other words,

I_n(s) = n/s I_{n-1}(s).

Observe that

I_n(s) = (n/s)((n-1)/s) I_{n-2}(s)

= (n/s)((n-1)/s)((n-2)/s)I_{n-3}(s)
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=(n/s)((n-1)/s)((n-2)/s)···(2/s)(1/s) I_0(s)

=n!/s^n I_0(s).

Now I_0(s) = ∫[t = 0 to ∞] e^{-st} dt = 1/s since s>0, hence

I_n(s) = n!/s^{n + 1}.

use the product rule