If your teacher wants a more combinatorial proof, then $\binom{2n}{n}$ is the number of $n$-element subsets of a $2n$-element set. Similarly, $\binom{2n}{n-1}$ is the number of $(n-1)$-element subsets of a $2n$-element set.
So, number the elements of the $2n$-element set $1,2,\ldots, 2n$. Consider a pair, $(i,\{a_1,\ldots, a_n\})$ where $1\le i \le n$ and $\{a_1,\ldots, a_n\}$ is an $n$-element subset of the $2n$-element set where $a_1<a_2 <\cdots <a_n$. The product principle tells us there are $n\binom{2n}{n}$ such pairs.
Given any such pair, $(i,\{a_1,\ldots, a_n\})$, remove the $i$-th element from the set, so that $\{a_1,\ldots, a_{i-1},a_{i+1},\ldots, a_n\}$ remains. Consider the remaining elements of the $2n$-element set: $\{b_1,\ldots, b_{n+1}\}$ (there are $n+1$ elements not in the set after one element is removed). Let $k$ be the index of the removed element. This allows us to create a pair, $(k,\{a_1,\ldots, a_{i-1},a_{i+1},\ldots, a_n\})$ where $1\le k \le n+1$ and the set is an $(n-1)$-element subset of the $2n$-element set. Again, the product principle tells us there are $(n+1)\binom{2n}{n-1}$ such pairs.
Now, given any such pair consisting of a number from 1 to $n+1$ and an $(n-1)$-element subset of the $2n$-element set: $(k,\{a_1,\ldots, a_{n-1}\})$, consider the remaining $n+1$ elements of the $2n$-element set: $\{b_1,\ldots, b_{n+1}\}$ where $b_1 < \cdots < b_{n+1}$. Add $b_k$ to the $n-1$-element subset. Reindex the set so that $\{a_1^\prime, \ldots, a_n^\prime\}$ satisfies $a_1^\prime < \cdots < a_n^\prime$. Let $i$ be the index of $b_k$ in this new index. This allows us to create a pair, $(i,\{a_1^\prime,\ldots, a_n^\prime\})$. These are inverse operations, implying a bijective process. If there is a bijection between two sets, they must have the same number of elements, so $n\binom{2n}{n} = (n+1)\binom{2n}{n-1}$.
The proof is a bit longer, but it also a bit stronger combinatorially.