Cases of Master Theorem
The three cases correspond exactly to the three cases in the statement of the theorem. Let's consider them one by one. Suppose for simplicity that $T(1) = 1$.

Case 1. Suppose that $f(n) = n^\delta$, where $\delta &lt; \log_b a$. Then
$$
T(n) = n^{\log_b a} + \sum_{i=0}^{\log_b n-1} \left(\frac{a}{b^\delta}\right)^i n^\delta.
$$
Since $\delta &lt; \log_b a$, we have $b^\delta &lt; a$ and so $a/b^\delta &gt; 1$. Therefore
$$
\sum_{i=0}^{\log_b n-1} \left(\frac{a}{b^\delta}\right)^i =
\left(\frac{a}{b^\delta}\right)^{\log_b n} \sum_{j=1}^{\log_b n} \left(\frac{b^\delta}{a}\right)^j = \Theta(n^{\log_b a - \delta}),
$$
since the geometric series $\sum_j (b^\delta/a)^j$ converges, and $\log_b (a/b^\delta) = \log_b a - \delta$. We conclude that
$$
T(n) = n^{\log_b a} + \Theta(n^{\log_b a - \delta} n^\delta) = \Theta(n^{\log_b a}).
$$
(So it is not quite true that the first term is dominant.)

Case 2. Suppose that $f(n) = n^{\log_b a}$ (this is the case $k = 0$; the case $k &gt; 0$ is similar though slightly more complicated). Then
$$
T(n) = n^{\log_b a} + \sum_{i=0}^{\log_b n-1} \left(\frac{a}{b^\delta}\right)^i n^{\log_b a}.
$$
This time $a = b^\delta$, and so each summand is equally dominant, leading to
$$
T(n) = n^{\log_b a} + (\log_b n) n^{\log_b a} = \Theta(n^{\log_b a} \log n).
$$

Case 3. Suppose that $f(n) = n^\delta$, where $\delta &gt; \log_b a$. Then
$$
T(n) = n^{\log_b a} + \sum_{i=0}^{\log_b n-1} \left(\frac{a}{b^\delta}\right)^i n^\delta.
$$
Since $\delta &gt; \log_b a$, we have $b^\delta &gt; a$, and so $a/b^\delta &lt; 1$. Therefore the geometric series $\sum_i (a/b^\delta)^i$ converges, implying that
$$
T(n) = n^{\log_b a} + \Theta(n^\delta) = \Theta(n^\delta).
$$



These proof sketches serve to explain that proof hints. However, they're not a complete proof, for two reasons. First, I assumed that $f$ has a specified form rather than a specified order of magnitude. Second, there is a tacit assumption that $n$ is a power of $b$. With more work we can take these points into account and prove the complete theorem. (A third and more minor point is the base case – you have assumed a base case of $1$, whereas the theorem calls for an arbitrary base case.)