The point is the following:
Delta, $\Delta$, is defined as $\frac{\partial C}{\partial S}$, where $C$ is the value of the call option, and $S$ is the price of the underlying asset.
So, given that the value of a call option for a non-dividend-paying underlying stock in terms of the Black–Scholes parameters is
$$C = N(d_{1})S - N(d_{2})Ke^{-rT},$$
$$\Delta = \frac{\partial C}{\partial S} = N(d_{1}).$$
Basically, Delta is just the first partial derivative of $C$ with respect to $S$.
How to derive $\Delta$
$N'(x)$ is the PDF for a standardized normal distribution:
- $N(x)$ is the cumulative probability that a variable with a standardized normal distribution will be less than x;
- $N'(x)$ is the probability density function for a standardized normal distribution:
$$N'(X) = \frac{1}{\sqrt{2\pi}}e^{\frac{x^2}{2}}.$$
Then, defining $\tau = T - t$, we have $$ d_{1} = \frac{\ln(\frac{S}{K}) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$$
and
$$ d_{2} = \frac{\ln(\frac{S}{K}) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$$
It follows that
$$ N'(d_{1}) = N'(d_{2} + \sigma\sqrt{\tau}) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(d_{2} + \sigma\sqrt{\tau})^2}{2}} = N'(d_{2})e^{-d_{2}\sigma\sqrt{\tau} - \frac{\sigma^2\tau}{2}} = N'(d_{2})\frac{Ke^{-r\tau}}{S}$$
Thus,
$$N'(d_{1})S = N'(d_{2})Ke^{-r\tau}.$$
Then
$$ \frac{\partial d_{1}}{\partial S} = \frac{\partial d_{2}}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}$$
Since there is an $S$ in $N(d_{1})$ and $N(d_{2})$, we use the chain-rule:
$$ \frac{\partial C}{\partial S} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{2})Ke^{-r\tau} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{1})S = N(d_{1}) + \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S - \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S = N(d_{1}).$$