# Graph of size n with girth >2k and minimum degree more than n^1/k

I'm stuck at a proof about Graphs in the context of Graph Spanners. A Lemma says:
Let $$G$$ be a graph with $$n$$ vertices and $$m$$ edges. If $$G$$ has girth more than $$2k$$, then $$m\leq n^{1+\frac{1}{k}}$$.
The proof is made by contradiction where repeatedly any node from $$G$$ of degree at most $$\lceil n^\frac{1}{k}\rceil$$ and any edges incident to that node is removed, until there is no such node anymore. This step leads to a subgraph $$G'$$ of $$G$$ of minimum degree more than $$\lceil n^\frac{1}{k}\rceil$$.
The conclusion is that $$G'$$ must have girth at most $$2k$$ and therefore also $$G$$, which is a contradiction.

I don't understand the conclusion: Why must $$G'$$ have girth at most $$2k$$? Why is it not possible that $$G$$ has girth more than $$2k$$ and minimum degree more than $$\lceil n^\frac{1}{k}\rceil$$?

I don't really see an answer to this question, so I would be very grateful if someone can help me.

Suppose that $$G'$$ has girth more than $$2k$$. Pick an arbitrary node $$v$$. The node $$v$$ has $$N_1 > \lceil n^{1/k} \rceil$$ neighbors. Each neighbor of $$v$$ has more than $$\lceil n^{1/k} \rceil$$ neighbors, and at least $$\lceil n^{1/k} \rceil$$ of them are different from $$v$$. The neighbors of different neighbors are all different, since otherwise $$G'$$ would have a 4-cycle. Denoting by $$N_2$$ the number of nodes at distance 2 from $$v$$, we have $$N_2 \geq \lceil n^{1/k} \rceil N_1$$. Continuing this way, we have $$N_\ell \geq \lceil n^{1/k} \rceil N_{\ell-1}$$ for all $$\ell \leq k$$ (here you use the lower bound on the girth; details left to you). In total, we get $$N_k > \lceil n^{1/k} \rceil^k \geq n$$, which is impossible.