Remember that $\displaystyle y_n = \int{e^{-t}t^{n}\,dt}$.
You have shown $\displaystyle \int{e^{-t}t^n\,dt} = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$.
So $\displaystyle y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}$.
Can you see that $\displaystyle \int{e^{-t}t^{n-1}\,dt}$ is almost the same as your original integral - the only difference being that $\displaystyle n$ is now $\displaystyle n-1$.
Surely this is $\displaystyle y_{n-1}$.
So what do you think $\displaystyle y_n$ is in terms of $\displaystyle y_{n-1}$?
That "log" is the natural logarithm and log(e)= 1!
I suspect you looked up the integral of $\displaystyle a^x$. You should know that $\displaystyle \int e^x dx= e^x+ C$ without having to look it up. If you let u= -x, then du= -dx so that $\displaystyle \int e^{-x}dx= -\int e^u du= -e^u+ C= -e^{x}+ C$.
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
Now, of course, state $\displaystyle y_{k}$ in terms of $\displaystyle y_{k-1}$ and k.
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
Can you at least see the pattern in
$\displaystyle y_{1} = y_{0} - e^{-1}$
$\displaystyle y_{2} = 2y_{1} - e^{-1}$
$\displaystyle y_{3} = 3y_{2} - e^{-1}$
and write it as
$\displaystyle y_{k} = $ ... ?
(or $\displaystyle y_{n} = $ ... ?)
I don't think any more is required than that. (And the starting value, for which, see above.)
No. For what seems like the one millionth time: $\displaystyle y_n$ is a DEFINITE integral!
$\displaystyle \displaystyle y_n = \left[-t^ne^{-t}\right]_{0}^{1} + n \int_0^1 {e^{-t}t^{n-1}\,dt}$.
Now the integral on the right hand side can be replaced with $\displaystyle y_{n-1}$. The first term on the right hand side needs to be evaluated in the usual way.