See Markov chain, Fundamental Theorem of Regular Markov Chains

Proof

a)

Let $W=\left(\begin{array}{c}\overrightarrow{w}\\ \overrightarrow{w}\\ ⋮\\ \overrightarrow{w}\end{array}\right)$, where $\overrightarrow{w}$ is a row vector probability distribution.
By definition of Markov Chain, $\underset{n\to \mathrm{\infty }}{lim}{P}^n=(\underset{n\to \mathrm{\infty }}{lim}{P}^{n-1})P$
So $W=WP$
Then $\left(\begin{array}{c}\overrightarrow{w}\\ \overrightarrow{w}\\ ⋮\\ \overrightarrow{w}\end{array}\right)=\left(\begin{array}{c}\overrightarrow{w}\\ \overrightarrow{w}\\ ⋮\\ \overrightarrow{w}\end{array}\right)P$

b)

Suppose $\overrightarrow{v}$ is a stationary distribution of $P$.
Then $\overrightarrow{v}=\overrightarrow{v}P=\overrightarrow{v}{P}^2=\dots =\overrightarrow{v}W$;
$\overrightarrow{v}=\overrightarrow{v}W=\left(\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right)\left(\begin{array}{c}\overrightarrow{w}\\ \overrightarrow{w}\\ ⋮\\ \overrightarrow{w}\end{array}\right)=(v_1+v_2+\dots +v_n)\overrightarrow{w}$

Since $\overrightarrow{v}$ is a probability distribution, $v_1+v_2+\dots +v_n=1$
So $\overrightarrow{v}=\overrightarrow{w}$.

QED