The adjusted R^2 formula is :
$$ \overline{R}^{2}=1-\left( \left( 1-R^{2}\right) \cdot > \dfrac{n-1}{n-k}\right) $$
In case of k > 1 , I continue like that;
$$ \overline{R}^{2}=1-\left( \dfrac{n-1}{n-k}-\dfrac{n-1}{n-k}R^{2}\right) $$
then
$$ \left( n-k\right) \cdot \overline{R}^{2}=n-k-\left( n-1\right) +\left( n-1\right) R^{2} $$
so $$ \left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) R^{2}=1-k $$
but after that I don't know how to proceed, is there someone who has an idea?
$$ SSRes=\sum_{i=1}^n\left( y_i-\hat y_i \right)^2\\ SSTotal=\sum_{i=1}^n\left( y_i-\bar y \right)^2 $$
$$ R^2=1-\dfrac{ SSRes/(n-1) }{ SSTotal/(n-1) } $$
$$ \bar R^2=1-\dfrac{ SSRes/(n-k) }{ SSTotal/(n-1) } $$
When $k>1$, the numerator of the second equation will be larger, meaning that the second equation subtracts a larger number from $1$ than does the first equation.
Thus, $R^2\ge \bar R^2.$ $\square$
A way to think about $R^2$ is that it compares the error variance to the total variance, and adjusted $\bar R^2$ does the same but with an estimate of the error variance that is slightly larger. Since that estimate of the error variance is slightly larger, adjusted $\bar R^2$ must be slightly less favorable (slightly smaller).
That's a great answer from Dave. I have another answer.
There is an alternative form of writting $\bar{R}^2$ in terms of $R^2$, with $p$ being the number of covariates in the model and $n$ the number of observations: $$ \bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1} $$ Now, subtract $p$ and add $p$ inside $(n-1)$: $$ \bar{R}^2 = \frac{(n-1)R^2 - p}{n-p-1} = \frac{(n-p-1+p)R^2 - p}{n-p-1} = \frac{(n-p-1)R^2 + pR^2 - p}{n-p-1} = \frac{(n-p-1)R^2 - p(1-R^2)}{n-p-1} = R^2 - \frac{p(1-R^2)}{n-p-1} $$
We can see that $\bar{R}^2$ is $R^2$ minus something nonnegative whenever $n > p + 1$, so $\bar{R}^2$ will be less or equal to $R^2$. For the "equal" part, $\bar{R}^2 = 1$ only when $R^2 = 1$ (knowing that $0 \le R^2 \le 1$).
Continuing from \begin{align*} \left( n-k\right) \cdot \overline{R}^{2}-\left( n-1\right) \cdot R^{2} & = 1-k \\ \left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot 1 & = \left( n-1\right) \cdot R^{2}. \end{align*} So $R^{2}$ is a weighted average of $\overline{R}^{2}$ and $1$, and since $\overline{R}^{2} \leq 1$: \begin{align*} \left( n-k\right) \cdot \overline{R}^{2} + (k-1) \cdot \overline{R}^{2} & \leq \left( n-1\right) \cdot R^{2} \\ \left(n-1\right) \cdot \overline{R}^{2} & \leq \left( n-1\right) \cdot R^{2} \\ \overline{R}^{2} & \leq R^{2} \\ \end{align*}
