Find $\int_0^1(\ln x)^n\hspace{1mm}dx$ I am not a big fan of induction, it's just a personal preference.
Is there a method other than induction.
Answer is $n!$ by the way 
 A: Given

$$
\int_0^1 \ln^n(x) dx
$$

Let us write
$$
I_n = \int_0^1 \ln^n(x) dx.
$$
Integrate by parts
$$
\begin{eqnarray}
I_n &=& \int_0^1 \ln^n(x) dx\\
&=& \Bigg[ x \ln^n(x) \Bigg]_0^1
      - n \int_0^1 \ln^{n-1}(x) dx\\
&=& \Bigg[ \exp(y) y^n\Bigg]_{-\infty}^0
      - n \int_0^1 \ln^{n-1}(x) dx
= - n I_{n-1}.
\end{eqnarray}
$$
Whence

$$
\bbox[16px,border:2px solid #800000] { I_n = - n I_{n-1} }
$$

Note that
$$
I_0 = \int_0^1 dx = 1.
$$
So the final result is

$$
\bbox[16px,border:2px solid #800000] {
  \int_0^1 \ln^n(x) dx = (-1)^n n! }
$$

A: Hint: Take $I(k)=\displaystyle\int_0^1x^k~dx,~$ and evaluate $I^{(n)}(0)$.
A: Outline of Method


*

*Perform the substitution $u = \ln x$. You will get
$$ \int_{-\infty}^0 x^n e^x dx.$$

*This is very nearly the Gamma function. Perform the substitution $u = -x$. You will get
$$ (-1)^n \int_0^\infty x^n e^{-x} dx.$$

*This now is the Gamma function. In particular, you have
$$ (-1)^n \Gamma(n+1) = (-1)^n n!.$$

*We might do a sanity check when $n = 1$. In this case,
$$ \int_0^1 \ln x dx = \Big[x \ln x - x  \Big ]_0^1 = -1,$$
which checks out. $\diamondsuit$

A: Let, $\ln x=t\implies \frac{dx}{x}=dt\implies dx=e^tdt$, we have $$\int_{0}^{1}(\ln x)^ndx=\int_{-\infty}^{0}(t)^ne^tdt$$ $$=\int_{0}^{\infty}(-t)^ne^{-t}dt$$ $$=(-1)^n\int_{0}^{\infty}e^{-t}t^ndt$$ Now, using Laplace transform $\color{blue}{\int_{0}^{\infty}e^{-st}f(t)dt=L[f(t)]}$ & $\color{blue}{L[t^n]=\frac{n!}{s^{n+1}}}$, we get  $$(-1)^n\int_{0}^{\infty}e^{-t}t^n dt=(-1)^n L[t^n]_{s=1}$$ $$=(-1)^n\left[\frac{n!}{s^{n+1}}\right]_{s=1}$$ $$=(-1)^n\left[\frac{n!}{(1)^{n+1}}\right]=(-1)^n(n!)$$ $$\implies \color{blue}{\int_{0}^{1}(\ln x)^ndx=(-1)^n(n!)}$$ Let, $n$ be an even integer then we have  $$ \color{blue}{\int_{0}^{1}(\ln x)^ndx=(-1)^n(n!)=n!}$$
