I'm given a discrete time signal which is as follows:

$$ x[n] = \alpha^n u[n],\\$$ $$ h[n] = u[n],\\$$

Why is unit function neglected in the following step:

$$ y[n] = \sum_{k=0}^{n} \alpha^{k}\\$$


It isn't. The convolution sum is


The step $u[k]$ in $(1)$ can be removed if the index $k$ starts at zero, and the step $u[n-k]$ can be removed by replacing the upper summation limit by $n$. For clarity, one should multiply the result by $u[n]$ because if $n<0$ the result of the convolution is zero:



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