How come the Bernstein operator creates a polynomial of the same degree as its input function?
The coefficients of $t^k$ for $k \gt m$ can cancel out.

For instance, let $f(x) = x^m$ and consider the coefficient of $t^n$ for $n \gt m$.

The coefficient we get is

$$n^{-m}\sum_{i=0}^{n} \binom{n}{i} i^m (-1)^{n-i}$$

which can be shown to be zero by starting with

$$\sum_{i=0}^{n} \binom{n}{i} x^{i} (-1)^{n-i} = (x-1)^{n}$$

And differentiating and multiplying by $x$, $m$ times, and then setting $x=1$.

For $m \lt n$, this indeed comes out to be zero.