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EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

EDIT 2: in fact Assumption 1 below implies Assumption 2 by the question's author's work, so by the arguments below Assumption 1 is equivalent to $\mathrm{Gal}(C_n)=S_2\wr S_n$. It appears likely that it holds for all $n$, but I don't have a proof.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ is odd (and the same for $(2n)!\cdot\mathrm{disc}(F_n)$).

Regarding Assumption 2, it should be possible to prove using a standard Newton polygon argument combined with some group theory (extensions of Jordan's theorem which are much less elementary) and Assumption 1 to rule out $A_n$.

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

EDIT 2: in fact Assumption 1 below implies Assumption 2 by the question's author's work, so by the arguments below Assumption 1 is equivalent to $\mathrm{Gal}(C_n)=S_2\wr S_n$. It appears likely that it holds for all $n$, but I don't have a proof.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ is odd (and the same for $(2n)!\cdot\mathrm{disc}(F_n)$).

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ is odd.

Regarding Assumption 2, it should be possible to prove using a standard Newton polygon argument combined with some group theory (extensions of Jordan's theorem which are much less elementary) and Assumption 1 to rule out $A_n$.

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

EDIT 2: in fact Assumption 1 below implies Assumption 2 by the question's author's work, so by the arguments below Assumption 1 is equivalent to $\mathrm{Gal}(C_n)=S_2\wr S_n$. It appears likely that it holds for all $n$, but I don't have a proof.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ is odd (and the same for $(2n)!\cdot\mathrm{disc}(F_n)$).

Regarding Assumption 2, it should be possible to prove using a standard Newton polygon argument combined with some group theory (extensions of Jordan's theorem which are much less elementary) and Assumption 1 to rule out $A_n$.

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ is odd.

Regarding Assumption 2, it should be possible to prove using a standard Newton polygon argument combined with some group theory (extensions of Jordan's theorem which are much less elementary) and Assumption 1 to rule out $A_n$.

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={n!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=n!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).

EDIT: I initially misread the definition of $C_n$ to be $\sum_{k=0}^n\frac {x^{2k}}{k!}$ and answered the wrong question. The general approach should still apply, but additional ideas are required for a complete solution. I edited the answer to match the original definition of $C_n$.

Denote $F_n=\sum_{k=0}^n\frac {x^{k}}{(2k)!}, C_n=\sum_{k=0}^n\frac{x^{2k}}{(2k)!}=F_n(x^2)$. I will sketch a proof of $\mathrm{Gal}(C_n)=S_2\wr S_n$ under the following two assumptions on $n$:

Assumption 1: $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ are not squares in $\mathbb Q$.

Assumption 2: $\mathrm{Gal}(F_n)=S_n.$

Which values of $n$ satisfy Assumption 1 should be determined separately, either by somehow computing $\mathrm{disc}(F_n)$ explicitly or by using Newton polygons and/or ramification to show that for some prime $p$ its exponent in $\mathrm{disc}(F_n)$ and $(2n)!\cdot\mathrm{disc}(F_n)$ is odd.

Regarding Assumption 2, it should be possible to prove using a standard Newton polygon argument combined with some group theory (extensions of Jordan's theorem which are much less elementary) and Assumption 1 to rule out $A_n$.

Henceforth I assume Assumptions 1 + 2.

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $F_n$ and $K=\mathbb Q(\alpha_1,\ldots,\alpha_n)$. The splitting field of $C_n$ is $L=\mathbb Q(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})=K(\sqrt{\alpha_1},\ldots,\sqrt{\alpha_n})$. There is a surjection $\mathrm{Gal}(C_n)\to\mathrm{Gal}(F_n)=S_n$ and it is enough to show that its kernel, which can be identified with $\mathrm{Gal}(L/K)$, is $(S_2)^n$. By Kummer theory this is equivalent to

Claim 1: $\prod_{i\in I}\alpha_i\not\in (K^\times)^2$ for any $\emptyset\neq I\subset\{1,\ldots,n\}$.

It follows from Assumption 1 that $\alpha_1\cdots\alpha_n={(2n)!}$ is not a square in $K$. Since $\mathrm{Gal}(E_n)=S_n$ is transitive on the $\alpha_i$, if $\alpha_1$ is a square in $K$ then so are $\alpha_2,\ldots,\alpha_n$ and then $\alpha_1\cdots\alpha_n$ is a square, a contradiction to our observation above. Similarly $\alpha_i\not\in (K^{\times})^2$ for any $i$.

Next consider $\alpha_1\alpha_2$. If it is a square, by the 2-transitivity of $\mathrm{Gal}(F_n)=S_n$ it follows that $\alpha_{2k-1}\alpha_{2k}$ is a square for any $k$ and therefore if $n$ is even $\alpha_1\cdots\alpha_n$ is a square (recall that $n$ is even), a contradiction to our observation above. If $n$ is odd then the above argument shows that $\alpha_1\cdots\alpha_n\alpha_1=(2n)!\cdot\alpha_1$ is a square, but this implies that $\alpha_1$ itself is a square (this follows from the fact that if $n<p<2n$ is a prime and $P_1,\ldots,P_\nu$ are the primes of $K$ lying over $p$ then $\alpha_1$ is divisible by some but not all of them by a Newton polygon argument).

Similarly $\alpha_i\alpha_j$ is not a square for any $i\neq j$.

Finally let $I\subset\{1,\ldots,n\}$ with $2<|I|<n$. Since $\mathrm{Gal}(F_n)=S_n$ is transitive on subsets of the same size we may assume WLOG that $I=\{1,\ldots,m\}$. If $\alpha_1\cdots\alpha_m$ is a square then so is $\alpha_1\cdots\alpha_{m-1}\alpha_{m+1}$ and therefore so is $\alpha_m\alpha_{m+1}$, contradicting what we have shown above. This completes the proof of Claim 1 in all cases (under Assumption 1).