At least n Spades in $14$ cards Standard $52$-card deck. $14$ randomly chosen cards.
Total number of Combinations: $C_{52}^{14}$.
The question is: what is the probability of having at least $n$ Spades with the dealt $14$ cards?
When $n=1$, my reasoning is to calculate the $p$ of missing the spades altogether and subtract the result from $1$, so the answer would be:
$$p(S>=1)=1-\frac{C_{39}^{14}}{C_{52}^{14}}$$
When $n=2$, can I use the same method "allowing" just one spade to appear in missing combinations, thus:
$$p(S>=2)=1-\frac{C_{40}^{14}}{C_{52}^{14}}$$
If I use the same logic for "at least 10 spades":
$$p(S>=10)=1-\frac{C_{48}^{14}}{C_{52}^{14}}$$
However, calculations showed me the unbelievably high result $(0.73)$ so I am sensing a mistake somewhere, please check my method and advise.
Thanks to all contributing to my answers, I am still earning my privilege to rate your posts.

You correctly determined the total number of ways that you can select $14$ cards from a deck of $52$ cards:
$$\mathcal{N}=\binom{52}{14}=1768966344600$$
We can in general calculate the number of ways that you can select $n$ spades and $14-n$ non-spades as follows:
$$N_{n}=\binom{13}{n}\binom{39}{14-n}$$
So, the number of ways to get combinations of cards such that there is "at least" $n$ spades, where $n \in \{0,\dots,13\}$ is:
$$N=\sum_{k=n}^{13}\binom{13}{k}\binom{39}{14-k}$$
So if we want at least $10$ spades then:
$$N=\sum_{k=10}^{13}\binom{13}{k}\binom{39}{14-k}=24246300$$
So we have the probability of selecting one of these combinations:
$$P=\frac{N}{\mathcal{N}}=\frac{24246300}{1768966344600} = 1.37065 \times 10^{-5}$$
