There are at least two ways of seeing this.
The urn interpretation of the distribution can be shown to be
The beta-binomial distribution can also be motivated via an urn model for positive integer values of $\alpha$ and $\beta$, known as the Polya urn model. Specifically, imagine an urn containing $\alpha$ red balls and $\beta$ black balls, where random draws are made. If a red ball is observed, then two red balls are returned to the urn. Likewise, if a black ball is drawn, then two black balls are returned to the urn. If this is repeated $n$ times, then the probability of observing k red balls follows a beta-binomial distribution with parameters $n$,$\alpha$ and $\beta$.
However if $n$ is negligible compared to the number of balls in the urn, adding a few balls back to the urns makes negligible difference for the next draw. It follows that the distribution is simply that of drawing with replacement, which is binomial.
From an algebraic viewpoint, the distribution is
$$
{n \choose k} \frac{B(k + \alpha, n - k + \beta)}{B(\alpha, \beta)}
.
$$
By the properties of the Beta function
$$
B(x + 1, y) = B(x, y) \frac{x}{x + y},
\\
B(x, y + 1) = B(x, y) \frac{y}{x + y}
$$
Specifically,
$$
B(i + \alpha, n - k + \beta) = B(i - 1 + \alpha, n - k + \beta) \frac{i - 1 + \alpha}{i - 1 + n - k + \alpha + \beta},
$$
and for large $\alpha, \beta$, taking into account the Taylor series of $\frac{1}{1 + x}$:
$$
\frac{i - 1 + \alpha}{i - 1 + n - k + \alpha + \beta} =
\frac{i - 1 + \alpha}{\alpha} \frac{\alpha}{(\alpha + \beta) \left(1 + \frac{i - 1 + n - k}{\alpha + \beta}\right)}
\sim
\frac{i - 1 + \alpha}{\alpha} \frac{\alpha}{(\alpha + \beta)} \left(1 - \frac{i - 1 + n - k}{\alpha + \beta}\right) \sim \frac{\alpha}{(\alpha + \beta)}
.
$$
Continuing this,
$$
\frac{B(k + \alpha, n - k + \beta)}{B(\alpha, \beta)}
\sim
\frac{B(\alpha, \beta)}{B(\alpha, \beta)} \left( \frac{\alpha}{\alpha + \beta}\right)^k \left( \frac{\beta}{\alpha + \beta} \right)^{n - k}
,
$$
and the distribution is approximately binomial.
