I see a proof of Sylow's first theorem (that Sylow subgroups exists) here. I change the proof a little bit when I try to reproduce it, but I am not sure if my proof is correct.
Theorem. Let $G$ be a group. $|G|=mp^n,p\nmid m $. ("$|$" means "divides".) Then $\text{Syl}_p(G)\neq \emptyset$.
My proof. Let's proceed by induction on the order of the group.
If $|Z(G)|=m_1p^k$, where $k\geq 1$ and $m_1 |m$. By induction hypothesis, $Z(G)$ has a sylow $p$-subgroup of order $p^k$, namely $P$. $Z(G)$ is Abelian, so the choice of $P$ is unique. (I admit we do need to prove some results for Abelian group first.) Therefore, $Z(G) \text{ char }G,P \text{ char }Z(G)$, and due to transitivity of $\text{ char }$, $P \text{ char } G$. Therefore, we can take the quotient group $G/P$. $|G/P|=m_2p^{n-k}$, where $m_1m_2=m$, by Lagrange theorem. By induction hypothesis, $G/P$ has a Sylow $p$-subgroup $P_2/P$ of order $p^{n-k}$.
Now we prove that $P_2\in \text{Syl}_p(G)$. $\forall x\in P_2,\exists r_1, x^{p^{r_1}}\in P$ since $P_2/P$ is a $p$-group. However, $P$ is also a $p$-group, so $\exists r_2, (x^{p^{r_1}})^{p^{r_2}}=x^{p^{r_1+r_2}}=1$, so $P_2$ is also a $p$-group. It's obvious that $|P_2|=p^n$.
The case in which $|Z(G)|$ is not divisible by $p$: I will prove that there exists a proper subgroup whose order is divisible by $p^n$, and apply the induction hypothesis. Suppose not. Then $\forall g\in G$ ,$|N_G(g)|$ is not divisible by $p^n$. Therefore, the length of the conjugate class of $g$, $|G|/|N_G(g)|$ is divisible by $p$. Let $\{C_i\}$ be the set of conjugate classes with length at least $2$. Then $$ |G|=|Z(G)|+\sum |C_i|, $$ which forces $|Z(G)|$ to be divisible by $p$, contradiction.
Is my proof correct? It does seem to be essentially the same as the one linked.
EDIT: $N_G(g)$ should really be $C_G(g)$.