On the Evolution of Random Graph (Existence of Log-size Components) Theorem:  Let $ \mathbf c > 0, p \geq \frac{c}{n}.$ Prove that there exists $a>0$ such that with asymptotical probability 1, the size of the largest component of $G(n,p)$ is at least $a\ln n$.
What I tried: By Cayley's formula I know that there are exactly $k^{k-2}$ spanning trees on $k$ vertices. And let $X_k = $ number of tree components of size $k$ in the graph $G(n,p)$. And the Expectation can be expressd as: $$E(X_k) = \binom{n}{k} k^{k-2}p^{k-1}(1-p)^{k(n-k)} \approx \frac{n^k e^k}{\sqrt{2\pi k}k^k}k^{k-1}(1-\frac{c}{n})^{k(n-k)} \approx \frac{n e^k c^{k-1}}{\sqrt{2\pi}k^{2.5}}(1-\frac{c}{n})^{kn} \approx \frac{n(e^{1-c}c)^k}{\sqrt{2\pi}k^{2.5}}$$
But I am stuck here. How do I prove this theorem? Any help would be appreciated.
 A: Thanks for the kind assistance and recommendations dear Suman Chakraborty and Misha Lavrov. Here is my complete attempt. Please help me with the evaluation
By Cayley's Formula we know that there are exactly $k^{k-2}$ spanning trees on $k$ vertices. let $p=\frac{c}{n}$
Let $X_k =$ number of tree components of size $k$ in a graph $G(n,p)$ and let $k = a \ln n$
$$ E(X_k) = \binom{n}{k} k^{k-2} p^{k-1} (1-p)^{k(n-k)} \backsim \frac{n^k e^k}{\sqrt{2 \pi k}k^k} k^{k-2} {\backsim{c}{n}}^{k-1} \left( 1-\frac{c}{n} \right)^{k(n-k)} $$
$$ \backsim \frac{ne^k c^{k-1}}{\sqrt{2\pi}k^{2.5}} \left( 1 - \frac{c}{n} \right)^{kn}  \backsim \frac{ne^k c^{k-1}}{\sqrt{2\pi}k^{2.5}}e^{kn \left( \frac{-c}{n} + \mathcal{O}(\frac{1}{n^2}) \right)} \backsim  \frac{ne^k c^{k-1}e^{-ck}}{\sqrt{2\pi}k^{2.5}} \backsim  \frac{ne^{(1-c)k} c^k}{\sqrt{2\pi}ck^{2.5}}$$
Let us simplify the numerator
$$ \hspace{0.2cm} Let \hspace{0.2cm} (e^{1-c}c)^k \geq \frac{1}{n} \implies n \geq (\frac{1}{n} e^{c-1})^k \implies \ln n \geq k \ln (\frac{1}{c}e^{c-1})^k $$
$$ \ln n \geq k \ln (\ln \frac{1}{c} + x-1) \implies k \leq \frac{1}{\ln \frac{1}{c} + c-1} \ln n \implies k \leq a\ln n $$ $$ \therefore EX_k \rightarrow \infty, k = a\ln n $$
It remains to compute the variance
$$ Var X_k = EX_k + EX_k (X-1) - (EX-k)^2$$ $$ EX_k(X_k - 1) = \binom{n}{k} \binom{n-k}{k}k^{2(k+2)}p^{2(k-1)} \left(1-p \right)^{2(\binom{k}{2} - k +1)} \left( 1-p \right)^{2k(n-k)}\left( 1-p \right)^{k^2} $$
We can see that $$ EX_k(X_k -1) \backsim (EX_k)^2 \implies Var X_k = \mathcal{O} ((EX_k)^2) $$ Applying Chebychev's inequality we get:
$$ P(X_k =0) \leq \frac{VarX_k}{(EX_k)^2} = \circ (1) $$
