How come the Bernstein operator creates a polynomial of the same degree as its input function? I read that the Bernstein operator

$$
\mathfrak{B}_f(t) = \sum_{i=0}^n f\left(\frac{i}{n}\right) \; B^n_i(t)
$$

applied to a polynomial $f(x)$ of degree $m \leq n$ with the Bernstein polynomial

$$
B^n_i(t) = \binom{n}{i} \; t^i \; (1-t)^{n-i}
$$

returns a new polynomial also of degree $m$. In other words, the Bernstein operator does not increase its degree when applied to $f$.

How is this possible? Since we're only sampling $f$ at a certain interval, wouldn't the original degree of $f$ become irrelevant? How can the degree of the result be smaller than $n$, since we will have at least one $t^n$ in the expression resulting from one of the Bernstein polynomials?
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.