As you can see in the picture, they simply multiplied the input in the $n$ domain with the Transfer function in the $w$ domain, to get the output in the $n$ domain. That's what they did, right?
This doesn't make sense to me. The proper way of doing this is to compute discrete time fourier transform of the input, multiply it with the transform of the transfer function and then do inverse transform.
The reason given is that 'complex exponential are eigensignals'. I have a rough idea about eigenvectors (that they only get scaled on transformations). So I guess something like that is happening here. But the problem is that integral transforms involve change of variable instead of just scaling.
This comes from the definition of a transfer function. An LTI system $H$ is usually characterized by its impulse response, $h[n]$, but the question is, why?
As you know the impulse response is simply $H(\delta[n])$. In discrete time, we use the Kronecker deltas as our basis. For an LTI system, we can use the resolution of identity to find
$$H(x[n]) = H\left(\sum_{k\in\mathbb{Z}}x[k]\delta[n-k]\right) = \sum_{k\in\mathbb{Z}} x[k]H(\delta[n-k]) = \sum_{k\in\mathbb{Z}} x[k]h[n-k] \equiv (x\star h)[n]$$
But let's consider another basis, namely the complex exponential basis, $\{e^{j\omega n}\}$. These guys are the Fourier transform of the Kronecker delta basis, $\{\delta[n-k]\}$. Applying the system to each of these basis vectors individually:
$$H\left(e^{j\omega n}\right) = \sum_{k\in\mathbb{Z}} e^{j\omega k}h[n-k] = \sum_{k\in\mathbb{Z}} e^{j\omega (n-k)}h[k] = e^{j\omega n}\sum_{k\in\mathbb{Z}} e^{j\omega -k}h[k] \equiv H\left(e^{j\omega}\right)e^{j\omega n}$$
by recognizing that last equality as the definition of the Fourier transform. Now, for a fixed $\omega$, the transfer function $H\left(e^{j\omega}\right)$ is a constant. In other words the complex exponentials are eigenvectors of the LTI signal with eigenvalue $H\left(e^{j\omega}\right)$, so they have merit as an alternative basis.
Suppose we represented our signals in this basis:
$$x[n] = \sum_{k\in\mathbb{Z}}x_ke^{jk\omega_0n}$$
Then the system applied to the signal would be
$$H(x[n]) = H\left(\sum_{k\in\mathbb{Z}}x_ke^{jk\omega_0 n}\right) = \sum_{k\in\mathbb{Z}} x_kH(e^{jk\omega_0 n}) = \sum_{k\in\mathbb{Z}} x_kH(e^{jk\omega_0 })e^{jk\omega_0 n}$$
So if you can write your input signal in this basis very nicely, then it becomes easier to compute the output, rather than trying to find the impulse response and computing the convolution by hand.