A tree with a node of degree $k\geq 1$ has at least $k$ pendant (degree 1) nodes. **Proof**: The case $k-1$ is trivial. Suppose $k>1.$ Let $n=|V|$ and let $\{ d_{i} \}_{i=1}^{n}$ be the degree sequence of $T.$ Let $l$ be the number of pendant nodes in $T.$ WLOG let $1=d_{1}=d_{2}=\dots=d_{l}$and $d_{l+1}=k.$ Now $|E|\leq {n \choose 2}=\frac{n-1}{2}$ $\sum d_{i} \geq l+k +2 (n-l-1)$ so $l+k+2(n-l-1)\leq 2(n-1) \implies l \geq k.$