Find number of leaves of a tree (Proof) Problem:
Let $T$ be a tree that has $i\ge1$ branch nodes, all of which have the same degree[1] $d$. Show that the number of leaves $(l)$ of the tree can be calculated via the following formula:
$$l=(d-2)i+2$$
[1] The degree of a node is defined as the number of all edges connected to it $(d_{leaf}=1)$.

What I've tried:
I've tried to solve this problem using induction to the number of leaves (I'm open to a better way).
Initial Notes:


*

*For a node to be a branch node: $d\ge2 \implies d_{min}=2$.

*As a result, the minimum possible number of leaves $l_0=2$.

*Also, let $s=d-d_{min}\ge0$.


So, the formula becomes:
$$
l=(d-d_{min})i+d_{min}
$$
Base Case:
Show that the formula holds for the minimum number of branch nodes $(i=1)$:
$$
l_1=(d-d_{min})\cdot 1+d_{min}=d
$$
The above holds, because the only branch node will be connected to all leaves. $l_1$ can also be written in the following form:
$$
\begin{align}
l_1
&=(d-d_{min})\cdot 1+d_{min}\\
&=l_0+(d-d_{min})\\
&=l_0+s
\end{align}
$$
Inductive Step:
Assume that the formula works for $(i=k)$:
$$
l_k=(d-d_{min})k+d_{min}
$$
Then prove that it works for $(i=k+1)$:
$$
\begin{align}
l_{k+1}
&=(d-d_{min})(k+1)+d_{min}\\
&=(d-d_{min})k+(d-d_{min})\cdot1+d_{min}\\
&=l_k+(d-d_{min})\\
&=l_k+s
\end{align}
$$
Therefore, the formula holds $\forall\ i\ge1$.

Question:
Is my approach correct? If not, what is the best approach to solving this problem?
 A: There is a slick proof using the Euler characteristic formula: if a tree has finitely many nodes then the number of nodes minus the number of edges equals $1$.
You've used $i$ for the number of nodes of degree $d$; to improve visuals, I'll use a capital $I$. I'll also use $L$ for the number of leaves, so the total number of vertices of the tree is $V = I+L$. And let me use $E$ for the number of edges. The Euler characteristic equation becomes
$$V-E=1
$$
$$I+L-E=1
$$
which we can solve for
$$E = I+L-1
$$
Next we count in an another way. Each edge has two endpoints. Each leaf is counted exactly once in this manner, and each branch node is counted exactly $d$ times. Therefore 
$$2E = L + d I
$$
Now we eliminate $E$:
$$2(I+L-1)=L+dI
$$
and simplify
$$L = (d-2)I+2
$$ 
Regarding the Euler characteristic formula, the simplest proof I know is an induction proof that is kind of like what you have written. 
A: I'd say the simplest way is to recall that the sum of degrees of all vertices is twice the number of edges, which is $n-1=i+l-1$ for the tree in question. Thus,$$2(i+l-1)=l+di\\\implies l=2+(d-2)i$$
