We take a look at Laplace's equation $\Delta u=0, u:\mathbb{R}^n\rightarrow\mathbb{R}$ and want to look for explicit solutions, firstly, since Laplace's equation is invariant under rotations, for radial solutions, i.e. $r=|x|=(x_1^2+\dots+x_n^2)^{1/2}$. Let us try to find a solution $u$ of the form $u(x)=v(r)$, where $v$ is to be selected so that $\Delta u=0$ holds, if possible. Note that $\frac{\partial r}{\partial x_i}=\frac{x_i}{r} (x\neq0,i=1,...,n)$. We thus have $u_{x_i}=v'(r)\frac{x_i}{r},\;u_{x_ix_i}=v''(r)\frac{x_i^2}{r^2}+v'(r)(\frac{1}{r}-\frac{x_i^2}{r^3})$, for $i=1,...,n$, and so $\Delta u=v''(r)+\frac{n-1}{r}v'(r)$. Hence $\Delta u=0$ holds iff $v'' +\frac{n-1}{r}v'=0$. If $v'\neq0$, we deduce $\log(|v'|)'= \frac{v''}{v'}=\frac{1-n}{r}$ and hence $v'(r)=\frac{a}{r^{n-1}}$ for some constant $a$.
Now it is this last step I do not understand, why does $v'(r)=\frac{a}{r^{n-1}}$ for some $a$? I tried to calculate to log, or to rewrite the fraction $\frac{v''}{v'}$,but I couldn't figure it out. (I did manage to check all previous steps)
Since $$ \frac{d}{dr}(\log |v'|)=\frac{1-n}{r}, $$ integration w.r.t. $r$ gives $$ \log |v'|=\int\frac{1-n}{r}\, dr=(1-n)\log r+C=\log r^{1-n}+C. $$ Exponentiating now gives $$ |v'|=e^Cr^{1-n} $$ or $$ v'=ar^{1-n},\quad a=\pm e^C, $$ where $a$ is some nonzero constant.