= Solution
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 $\mathbb P_p(0\leftrightarrow\partial[-R,R]^2)\le Ce^{-cR}$ 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 gives
$$
\chi:=\mathbb E_p|C_0|=\sum_{v\in\mathbb Z^2}\tau(0,v)\le1+\sum_{R\ge1}8R\,Ce^{-cR}<\infty.
$$
A 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>:
$$
\mathbb E_p|C_0|^k\le (2k-3)!!\,\chi^{2k-1},\qquad k\ge1,
$$
where $(-1)!!=1$.
Expand the left-hand side as a sum over ordered marked vertices $x_1,\ldots,x_k$ of the <probability> that $0,x_1,\ldots,x_k$ 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 $(2k-3)!!\le2^{k-1}(k-1)!$, expansion of the exponential, justified by <monotone convergence>, gives
$$
\mathbb E_p e^{t|C_0|}\le1+\frac1{2\chi}\sum_{k\ge1}\frac{(2t\chi^2)^k}{k}=1-\frac1{2\chi}\log(1-2t\chi^2),\qquad 0<t<\frac1{2\chi^2}.
$$
Take $t=1/(4\chi^2)$ and write $M=1+\log2/(2\chi)$. <Markov's inequality> yields $\mathbb P_p(|C_0|\ge n)\le Me^{-tn}$. To remove the prefactor, choose an integer $N\ge2$ with $tN\ge2\log M$. For $n\ge N$ the bound is at most $e^{-tn/2}$. For $2\le n\le N$, use
$$
\mathbb P_p(|C_0|\ge n)\le\mathbb P_p(|C_0|\ge2)=1-(1-p)^4=:u<1.
$$
For $0<p<1/2$, setting $a=\min\{t/2,-\log u/N\}>0$ therefore proves \b[the bound with unit prefactor]:
$$
\boxed{\mathbb P_p(|C_0|\ge n)\le e^{-a n}\quad(n\ge2).}
$$
At $p=0$, $|C_0|=1$ and any positive $a$ works.
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