The current data point is said to have age 0, the previous has age 1, and so on going backwards.
For a straight N period moving average of the form $\frac{1}{N}(x_t+x_{t-1}+\cdots+x_{t-N+1})$ it is easy to see that the average age of the data is $\frac{N-1}{2}$. Sometimes this is stated in term of "centering": a 3 period moving average is centered on the period $t-1$, i.e the period with age $1=\frac{3-1}{2}$.
A slightly more elaborate calculation shows that for an exponential moving average with constant $\alpha$, the average age of the data is $\frac{1-\alpha}{\alpha}$. (The EMA is of the form $\sum_{k=0}^\infty\alpha(1-\alpha)^k x_{t-k}$ and the average age is $\sum_{k=0}^\infty\alpha(1-\alpha)^k k$ which can be show to converge to $\frac{1-\alpha}{\alpha}$).
To find the EMA most similar to a given MA, we set these two expressions for average age equal, giving the equation $\frac{N-1}{2}=\frac{1-\alpha}{\alpha}$. Solving this for alpha we get $\alpha=\frac{2}{N+1}$ . QED.
This proof was given by Brown in his 1963 book 'Smoothing, Forecasting and Prediction'