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We use three precise inputs. The Harris-Kesten theorem says that independent bond percolation on the square lattice has percolation critical probability $p_c=1/2$. Sharpness of the percolation transition says that, for every $p<p_c$, there are $c,C>0$ such that ${\mathbb{P}}_p(0\leftrightarrow \partial [-R,R{]}^2)\le C{e}^{-cR}$ for all positive integers $R$. Finally, the BK inequality says that increasing connection events with disjoint bond witnesses have joint probability at most the product of their individual probabilities. All three statements concern the independent product law used here.

The sharpness bound implies finite percolation susceptibility. Indeed, with $\tau (u,v)={\mathbb{P}}_p(u\leftrightarrow v)$, summing over square shells givesA radius bound alone would give only a square-root exponential volume bound. To obtain the requested exponential volume bound, we prove the tree-graph moment bound for a percolation cluster:where $(-1)!!=1$.

Expand the left-hand side as a sum over ordered marked vertices $x_1,\dots ,x_k$ of the probability that $0,x_1,\dots ,x_k$ belong to one open cluster. Whenever that event occurs, a finite open tree connects the marked vertices. Delete unmarked leaves and suppress unmarked degree-two vertices. Treat a marked internal vertex as a leaf attached by a zero-length connection, and split any higher-degree branch into degree-three branch points, again with zero-length connections. This gives an abstract binary tree with $k+1$ labelled leaves, $k-1$ branch points and $2k-1$ connections. Its nonempty connecting paths use disjoint sets of open bonds. Coincident marked vertices are allowed: their zero-length connections have empty witnesses.

There are $(2k-3)!!$ such binary tree topologies. For example, the topology with two leaves is unique; inserting the next labelled leaf subdivides one of the $2k-1$ edges of a topology with $k+1$ leaves, giving the usual double-factorial recursion. For a fixed topology and fixed positions of all its vertices, the BK inequality bounds the disjoint connection event by the product of the corresponding $\tau$ factors. Sum over all $x_i$ and all branch-point positions. Keep the root fixed, and successively sum a leaf away. Translation invariance gives ${\sum }_y\tau (z,y)=\chi$ at every step. The resulting sum is ${\chi }^{2k-1}$. Summing the topologies proves the bound.

Since $(2k-3)!!\le 2^{k-1}(k-1)!$, expansion of the exponential, justified by monotone convergence, givesTake $t=1/(4{\chi }^2)$ and write $M=1+\log 2/(2\chi )$. Markov's inequality yields ${\mathbb{P}}_p(|C_0|\ge n)\le M{e}^{-tn}$. To remove the prefactor, choose an integer $N\ge 2$ with $tN\ge 2\log M$. For $n\ge N$ the bound is at most $e^{-tn/2}$. For $2\le n\le N$, useFor $0<p<1/2$, setting $a=\min \{t/2,-\log u/N\}>0$ therefore proves the bound with unit prefactor:At $p=0$, $|C_0|=1$ and any positive $a$ works.