With a Bernoulli process success probability $p$ and failure probability $q = 1 - p$, the probability for $T_k$ time of the $k^{th}$ success is $$P(T_k \leq n) = \sum_{j=k}^n \binom{n}{j}p^jq^{n-j}$$
My question is how does this work for $k=1$? According to the binomial theorem, the probability would be 1 for any $n$, and this doesn't make sense to me. For example, consider $n=7$. What if the first seven steps are failures, and $T_1 = 8$? It wouldn't make sense that $P(T_1 \leq 7) = 1$.
No, $\sum_{j=1}^n \binom{n}{j} p^j q^{n-j} \ne 1$. You are conflating it with $\sum_{j=0}^n \binom{n}{j} p^j q^{n-j} = 1$.