Assuming that $a^m$ occurs in the $(r+1)^{\text{th}}$ term of $(1+a)^{m+n}$, $T_{r+1} = ^{m+n}\large C$$_r\;1^{m+n–r}a^r$.
Comparing the indices of $a \rightarrow r = m$
Therefore, the coefficient of $a^m = \;^{m+n}\large C$$_{m} = \large\frac{(m+n)!}{m!(m+n-m)!}$$ = \large\frac{(m+n)!}{m!n!}$
Assuming that $a^n$ occurs in the $(k+1)^{\text{th}}$ term of $(1+a)^{m+n}$, $T_{k+1} = ^{m+n}\large C$$_k\;1^{m+n–k}a^l$.
Comparing the indices of $a \rightarrow k = n$,
Therefore, the coefficient of $a^n = \;^{m+n}\large C$$_{n} = \large\frac{(m+n)!}{n!(m+n-n)!}$$ = \large\frac{(m+n)!}{n!m!}$
$\Rightarrow$ we can see that the two coefficients are equal.