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I am trying to show the following:

enter image description here

I think I have tried to show the second part of. But I couldn't show the first part.

My proof for the second part is as follows:

enter image description here

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    Good. Now, try to prove the converse by using the standard homomorphism $F(S)\to G$. – 2017-01-25
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    For 1, the bound you are trying to prove is the total number of reduced words of length lass than $n$. Since every element of the group within the open ball of radius $n$ has such an expression as a reduced word, that number is an upper bound on the total number of elements in the ball. – 2017-01-25
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    Thanks! Do you have some reference on this bound. I couldn't find anything on it in literature. – 2017-01-25

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You have shown that the free group attains this bound. You have two things left to prove:

a. That every group satisfies this bound. The hint in the comment is a nice way of doing this: consider the natural homomorphism $F(S)\rightarrow G$.

b. That no other group attains the bound. To do this, consider a non-empty word $W(S)$ which is equal to the identity in $G$ (e.g. $G\cong \langle S\mid A, B, ...\rangle$ and take $W(S)=A$). This corresponds to a loop in the Cayley graph. Can you see how the result follows?

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    Thanks for the response, but I still don't see it. How does the natural homomorphism imply that every group satisfies the bound? – 2017-01-25
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    @matan: If $f:B'\to B$ is a surjective map then $|B|\le |B'|$. The cardinalities are equal iff $f$ is bijective. – 2017-01-25
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    @MoisheCohen Thanks! I think I got it now. – 2017-01-25