Laplace transformation of e^-t if t≥1

Hello,

I have to calculate the laplace transformation of this function

f(t) =
0 if t<1
e^-t if t≥1

I know it is a shift in t-domain, so i have multiply it with e^-t
The laplace transform of e^-t is 1/(s+1)

is my F(s) = 1/(s+1) * e^-t?

Because if i use the formula to caculate it i get another answer

F(s) = [1 to infinity] ∫ e^-t * e^-st dt
F(s) = e^(-(1+s))/(s+1)

But what is the right answer?

Could someone please help me?

I use S for integral.
From the definition F(s)=S(e^(-st)f(t) dt) from t=0 to inf
but since f(t)=0 from t=0 to 1
F(s)=S(e^(-st)e^(-t)dt from t=1 to inf
=S(e^(-(1+s)t) dt fro t=1 to inf
=[-(1/(1+s))e^(-(1+s)t)] from t=1 to inf
=e^(-(1+s))/(s+1)

I think you should try first principles before trying to apply a rule.

Egwewgw